MATH 112 - TECHNICAL
MATHEMATICS - Applications to
systems of linear equations
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
4x−y+ 2z= 5
3x+ 2y−3z=−2
Solution
Step 1: Write the system of equations in matrix form.
2 3 −1
4−1 2
3 2 −3
x
y
z
=
1
5
−2
Step 2: Use row operations to put the augmented matrix into row-echelon
form.
2 3 −1|1
4−1 2 |5
3 2 −3| −2
2 3 −1|1
0−7 4 |3
0−1−6| −5
2 3 −1|1
0−7 4 |3
0 0 −29/7| −2
Step 3: Solve for zusing back substitution.
−29
7z=−2 =⇒z=14
29
Step 4: Substitute z=14
29 back into the second row to solve for y.
−7y+ 4 14
29= 3 =⇒y=−35
29
Step 5: Substitute z=14
29 and y=−35
29 back into the first row to solve for
x.
2x+ 3 −35
29−14
29 = 1 =⇒x=49
29
Therefore, the solution to the system of equations is:
x=49
29, y =−35
29, z =14
29
Question 2
Question
Solve the following system of linear equations:
2x−3y+z= 7
4x+y−z= 1
3x−2y+ 2z= 10
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |7
4 1 −1|1
3−2 2 |10
Step 2: Perform row operations to simplify the matrix. Start by multiplying
the first row by 2 and subtracting 4 times the first row from the second row,
and subtracting 3 times the first row from the third row to get:
2−3 1 |7
0 13 −5| −27
0 7 −1| −11
Step 3: Next, divide the second row by 13 to simplify the matrix further:
2−3 1 |7
0 1 −5
13 | −27
13
0 7 −1| −11
2
Step 4: Perform row operations to eliminate the 7 in the third row. Subtract
7 times the second row from the third row to get:
2−3 1 |7
0 1 −5
13 | −27
13
0 0 32
13 |10
Step 5: Divide the third row by 32
13 to find the value of z:
2−3 1 |7
0 1 −5
13 | −27
13
0 0 1 |130
32
Step 6: Substitute z=130
32 back into the second row to find the value of y:
1y−5
13z=−27
13y−5
13 ×130
32 =−27
13y−25
16 =−27
13y=−27
13 +25
16 =−79
26
Step 7: Finally, substitute z=130
32 and y=−79
26 back into the first row to
find the value of x:
2x−3y+z= 72x−3×−79
26+130
32 = 72x+237
26 +130
32 = 72x= 7−237
26 −130
32 2x=182
13 x=91
13
Therefore, the solution to the system of linear equations is x=91
13 ,y=−79
26 ,
and z=130
32 .
Question 3
Question
Solve the following system of equations:
2x−3y+z= 4
4x+y−2z=−3
x−2y+ 3z= 5
Solution
Step 1: Let’s write the system of equations in matrix form:
2−3 1
4 1 −2
1−2 3
x
y
z
=
4
−3
5
Step 2: We will perform row operations to solve this system of equations.
First, we will subtract the first row multiplied by 2 from the second row, and
subtract the first row multiplied by 1
2from the third row:
2−3 1
0 7 −4
0−5
2
5
2
x
y
z
=
4
−11
7
2
3
Step 3: Next, we will divide the second row by 7 and the third row by −5
2:
2−3 1
0 1 −4
7
0 1 −1
x
y
z
=
4
−11
7
−7
10
Step 4: Now, we’ll subtract the second row from the third row:
2−3 1
0 1 −4
7
0 0 −3
10
x
y
z
=
4
−11
7
1
10
Step 5: Lastly, we can back substitute to find the values of x,y, and z. We
get z=−1
3,y=−4
7, and x=11
3.
Therefore, the solution to the system of equations is x=11
3,y=−4
7, and
z=−1
3.
Question 4
Question
Solve the following system of linear equations:
2x−y+z= 3
x+ 3y−2z=−4
3x−2y+ 2z= 8
Solution
Step 1: We start by writing the system of linear equations in matrix form
Ax=b, where Ais the coefficient matrix, xis the column matrix of variables,
and bis the column matrix of constants.
2−1 1
1 3 −2
3−2 2
x
y
z
=
3
−4
8
Step 2: Next, we perform row operations to get the matrix in row-echelon
form:
2−1 1 |3
1 3 −2| −4
3−2 2 |8
Row 2 - 1
2Row 1
Row 3 - 3
2Row 1
2−1 1 |3
07
2−5
2| −11
2
01
2
1
2|11
2
4
Step 3: Continue row operations to simplify the matrix further.
Row 2 ×2
7
Row 3 - 1
2Row 2
2−1 1 |3
0 1 −1| −11
7
0 0 2 |30
7
Step 4: Finally, we convert the matrix back into equations and solve for x,
y, and z. From the last row, we have 2z=30
7, so z=15
7.
Substitute z=15
7back into the second row: y−15
7=−11
7, which gives
y= 2.
Now substitute y= 2 and z=15
7back into the first row: 2x−2 + 15
7= 3,
which yields x= 1.
Therefore, the solution to the system of linear equations is x= 1, y= 2, and
z=15
7.
Question 5
Question
Find the values of aand bsuch that the following system of linear equations
has infinitely many solutions:
2x−3y= 4
ax +by = 2
Solution
Step 1: To have infinitely many solutions, the two equations must be linearly
dependent; in other words, one equation must be a scalar multiple of the other.
This means that the ratios of the coefficients of the variables in the two equations
must be equal. Therefore, we need to find aand bsuch that:
2
a=−3
b=4
2
Step 2: From the first ratio, we have 2
a=4
2, which simplifies to 2a= 4.
Solving for a, we get a= 2.
Step 3: Now substitute a= 2 into the second ratio: −3
b=4
2, which simplifies
to −3b= 8. Solving for b, we get b=−8
3.
Step 4: Therefore, the values of aand bsuch that the system of linear
equations has infinitely many solutions are a= 2 and b=−8
3.
5
Question 6
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+y−z= 2
3x−4y+ 2z= 6
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 |4
1 1 −1|2
3−4 2 |6
Step 2: Perform row operations to simplify the matrix.
1 1 −1|2
0−5 3 |0
0−7 5 |0
Step 3: Perform further row operations to obtain the reduced row-echelon
form.
1 1 −1|2
0 1 −3
5|0
0 0 1 |0
Step 4: Back-substitute to find the values of z,y, and x. From the last
row, z= 0. Substituting z= 0 back into the second row yields:
y−3
5= 0 ⇒y=3
5
Substituting y=3
5and z= 0 into the first row gives:
x+3
5−1=2⇒x=7
5
Step 5: Therefore, the solution to the system of equations is:
x=7
5, y =3
5, z = 0
6
Question 7
Question
Solve the following system of linear equations using the method of your choice:
3x−2y+z= 7
2x+ 4y+z= 4
x−3y−z=−5
Solution
Step 1: Rewrite the system of equations in matrix form.
3−2 1
2 4 1
1−3−1
x
y
z
=
7
4
−5
Step 2: Perform row operations to simplify the matrix equation.
3−2 1 |7
2 4 1 |4
1−3−1| −5
Step 3: Perform row operations to get the matrix in row-echelon form.
1−3−1| −5
0 10 4 |19
0 1 4 |12
Step 4: Continue to row-reduce the matrix to get it in reduced row-echelon
form.
1 0 −7| −23
014/10 |19/10
0 0 1 |11
Step 5: Convert the reduced matrix back into a system of equations.
x=−23
y=19
10
z= 11
Therefore, the solution to the system of equations is x=−23, y=19
10 , and
z= 11.
7
Question 8
Question
Solve the following system of linear equations:
3x−2y+ 4z= 7
x+ 2y−z=−1
2x−3y+ 5z= 10
Solution
Step 1: To solve the system of equations, we will use the method of substitution.
First, we will solve the second equation for x:
x=−1 + z−2y
Step 2: Next, we will substitute xinto the first and third equations:
3(−1 + z−2y)−2y+ 4z= 7
2(−1 + z−2y)−3y+ 5z= 10
Step 3: Simplifying the equations, we have:
−3+3z−6y−2y+ 4z= 7
−2+2z−4y−3y+ 5z= 10
Step 4: Simplifying further, we get:
7y+ 7z= 10
7y+ 7z= 12
Step 5: It is clear that the system of equations is inconsistent, as the last
two equations are contradictory. Therefore, there is no solution to the system
of equations.
Question 9
Question
Solve the following system of linear equations using matrix methods:
2x−y+ 3z= 4
x+ 2y+z= 1
3x−2y+ 2z= 5
8
Solution
Step 1: Write the system of equations in matrix form as AX =B, where
A=
2−1 3
121
3−2 2
, X =
x
y
z
, B =
4
1
5
Step 2: Find the inverse of matrix Aby calculating |A|and A−1.
|A|= 2(2)(2)+1(−2)(3)+3(1)(−1)−3(2)(3)−1(−2)(2)−2(1)(2) = 4−6−3−18+4−4 = −23
A−1=1
|A|
213
2 2 −2
−1 1 −1
=
−2
23 −1
23 −3
23
−2
23 −2
23
2
23
1
23 −1
23
1
23
Step 3: Calculate X=A−1B
X=
−2
23 −1
23 −3
23
−2
23 −2
23
2
23
1
23 −1
23
1
23
4
1
5
=
−4
23 −1
23 −15
23
−8
23 −2
23 +10
23
4
23 +1
23 +5
23
=
−20
23
0
10
23
Therefore, the solution to the system of equations is x=−20
23 ,y= 0, and
z=10
23 .
Question 10
Question
Solve the following system of linear equations:
2x+y+z= 4
x−y+ 3z= 5
3x+ 2y−z= 1
Solution
Step 1: Rewrite the system of equations in matrix form.
2 1 1
1−1 3
3 2 −1
x
y
z
=
4
5
1
Step 2: Find the determinant of the coefficient matrix to determine if the
system has a unique solution.
det
2 1 1
1−1 3
3 2 −1
=−1−(−6) = 5 = 0
9
Since the determinant is nonzero, the system has a unique solution.
Step 3: Use Cramer’s Rule to find the solution.
x=
det
4 1 1
5−1 3
1 2 −1
det
2 1 1
1−1 3
3 2 −1
=31
5
y=
det
2 4 1
1 5 3
3 1 −1
det
2 1 1
1−1 3
3 2 −1
=17
5
z=
det
214
1−1 5
321
det
2 1 1
1−1 3
3 2 −1
=1
5
Therefore, the solution to the system of linear equations is x=31
5,y=17
5,
and z=1
5.
Question 11
Question
Solve the following system of linear equations using any method of your choice:
3x−2y+z= 4
x+ 3y−2z= 6
2x−y+ 3z= 8
Solution
Step 1: We can solve this system of linear equations using the method of sub-
stitution or elimination. Let’s use the elimination method.
Step 2: Start by eliminating the variable xfrom the system. Multiply the
first equation by 1, the second equation by 3, and leave the third equation as is:
3x−2y+z= 4
3x+ 9y−6z= 18
2x−y+ 3z= 8
10
Step 3: Subtract the first equation from the second equation to eliminate x:
11y−7z= 14
Step 4: Now, eliminate xfrom the first and third equations. Multiply the
first equation by 2, the third equation by 3, and leave the second equation as is:
6x−4y+ 2z= 8
6x−3y+ 9z= 24
2x−y+ 3z= 8
Step 5: Subtract the third equation from the first equation to eliminate x:
3y−7z= 0
Step 6: We now have a system of two equations with two variables:
(11y−7z= 14
3y−7z= 0
Step 7: Solve this system to find yand z. After finding the values of yand
z, substitute them back into any of the original equations to solve for x.
Step 8: Finally, write the solution as an ordered triple (x, y, z).
Question 12
Question
Solve the following system of linear equations using the Gaussian elimination
method:
2x+5y−z= 9
x+3y+2z= 17
3x+7y+4z= 25
Solution
Step 1: Write the augmented matrix for the system of equations.
2 5 −1|9
1 3 2 |17
3 7 4 |25
Step 2: Perform row operations to simplify the matrix.
2 5 −1|9
1 3 2 |17
3 7 4 |25
→
1 3 2 |17
2 5 −1|9
3 7 4 |25
11
Step 3: Perform row operations to simplify the matrix further.
1 3 2 |17
2 5 −1|9
3 7 4 |25
→
1 3 2 |17
0−1−5| −25
0−2−2| −34
Step 4: Perform row operations to get the matrix in row-echelon form.
1 3 2 |17
0−1−5| −25
0−2−2| −34
→
1 3 2 |17
0 1 5 |25
0−2−2| −34
Step 5: Perform row operations to obtain the reduced row-echelon form.
1 3 2 |17
0 1 5 |25
0−2−2| −34
→
1 0 −13 | −34
0 1 5 |25
0 0 8 |16
Step 6: Solve for the variables. From the row-echelon form, we have:
x=−34, y = 25, z = 2
Therefore, the solution to the system of equations is x=−34, y= 25, and
z= 2.
Question 13
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
x−2y+ 2z=−3
3x+ 2y−3z= 5
Solution
Step 1: We will solve the system of equations using the method of substitution.
Step 2: From the first equation, we can express zin terms of xand y:
z= 2x+ 3y−4
Step 3: Substitute the expression for zinto the second and third equations:
x−2y+ 2(2x+ 3y−4) = −3
3x+ 2y−3(2x+ 3y−4) = 5
12
Step 4: Simplify the equations:
x−2y+ 4x+ 6y−8 = −3
3x+ 2y−6x−9y+ 12 = 5
Step 5: Combine like terms:
5x+ 4y= 5
−3x−7y=−7
Step 6: Solve this system of equations. Multiplying the first equation by 3 and
the second equation by 5 to eliminate ygives:
15x+ 12y= 15
−15x−35y=−35
Step 7: Adding the two equations results in:
−23y=−20
Step 8: Solve for y:
y=20
23
Step 9: Substitute yback into the equation 5x+ 4y= 5 to solve for x:
5x+ 4 20
23= 5
5x+80
23 = 5
5x=115
23
x=23
23 = 1
Step 10: Substitute xand yback into the equation z= 2x+ 3y−4 to solve for
z:
z= 2(1) + 3 20
23−4 = 2 + 60
23 −92
23 =−30
23
Step 11: Therefore, the solution to the system of equations is x= 1, y=20
23 ,
and z=−30
23 .
Question 14
Question
Solve the following system of linear equations:
2x−3y+z= 1
x+ 4y−2z=−2
3x−2y+ 5z= 10
13
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−3 1 |1
1 4 −2| −2
3−2 5 |10
Step 2: Perform row operations to row-echelon form:
100|3
010| −1
001| −2
Step 3: The solution to the system is x= 3, y=−1, and z=−2.
Question 15
Question
Solve the following system of linear equations:
2x−y+ 3z= 5
x+y+ 2z= 4
3x+ 5y−z= 2
Solution
Step 1: Write the system of equations in matrix form. Let’s denote the coeffi-
cient matrix as A, the variable matrix as X, and the constant matrix as B. The
system AX =Bcan be represented as:
2−1 3
1 1 2
3 5 −1
x
y
z
=
5
4
2
Step 2: Use Gaussian elimination to row-reduce the augmented matrix [A|B].
2−1 3 |5
1 1 2 |4
3 5 −1|2
Step 3: Perform row operations to row-reduce the augmented matrix:
1 1 2 |4
0 3 −1|1
0 2 −7| −10
14
Step 4: Continue row operations:
1 1 2 |4
0 1 −1
3|1
3
0 0 −19
3| −11
3
Step 5: Write the system of equations back with row-reduced form:
x+y+ 2z= 4
y−1
3z=1
3
−19
3z=−11
3
Step 6: Solve for z:z=−11
19
Step 7: Substitute zback into the second equation to solve for y:y−1
3·
−11
19 =1
3
y+11
57 =19
57
y=8
19
Step 8: Substitute yand zback into the first equation to solve for x:x+
8
19 + 2 ·−11
19 = 4
x=35
19
Therefore, the solution to the system of equations is x=35
19 ,y=8
19 ,z=−11
19 .
Question 16
Question
Solve the following system of linear equations using matrix methods:
2x−3y+z= 5
x+y+ 2z= 6
3x−2y−3z=−7
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
1 1 2
3−2−3
x
y
z
=
5
6
−7
Step 2: Set up the augmented matrix and perform row operations.
[ccc|c]2 −3 1 5
1 1 2 6
3−2−3−7
15
Step 3: Perform row operations to get the matrix in row-echelon form.
[ccc|c]1 1 2 6
0−5−3−7
0 0 0 0
Step 4: Solve for the variables starting from the bottom row.
−5y−3z=−7 =⇒y= 1 −3
5z
Step 5: Substitute yinto the next equation and solve for x.
x+ 1 −3
5z+ 2z= 6 =⇒x= 2 + 8
5z
Step 6: Substitute xand yback into the first equation to find z.
2(2 + 8
5z)−3(1 −3
5z) + z= 5 =⇒z= 5
Step 7: Finally, substitute zback into the expressions for xand yto get
the solution.
x= 2 + 8
5(5) = 10, y = 1 −3
5(5) = −2
Therefore, the solution to the system of equations is x= 10, y=−2, and
z= 5.
Question 17
Question
Solve the following system of linear equations:
3x−2y+z= 5
x+ 4y−2z=−10
2x−y+ 3z= 4
Solution
Step 1: Write the system of equations in matrix form.
3−2 1
1 4 −2
2−1 3
x
y
z
=
5
−10
4
Step 2: Find the inverse of the matrix on the left side of the equation.
Let A=
3−2 1
1 4 −2
2−1 3
16
To find A−1, we will row reduce the augmented matrix [A — I].
3−2 1 |100
1 4 −2|010
2−1 3 |001
Step 3: Perform row operations to convert the left side to the identity matrix.
100|5
17 −3
17
1
17
010| − 2
17
1
17
1
17
001| − 6
17
2
17
1
17
Therefore, the inverse of matrix Ais:
A−1=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
Step 4: Calculate the solution vector by multiplying A−1with the column
vector on the right side of the equation.
x
y
z
=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
5
−10
4
Step 5: Perform matrix multiplication to find the solution vector.
x
y
z
=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
5
−10
4
=
7
17
−12
17
1
17
Therefore, the solution to the system of equations is x=7
17 ,y=−12
17 , and
z=1
17 .
Question 18
Question
Consider the following system of linear equations:
2x+ 3y+z= 6
x+y+ 2z= 4
3x+ 2y+ 2z= 10
Determine if the system of equations is consistent. If it is consistent, find
the solution to the system. If it is inconsistent, explain why.
17
Solution
Step 1: Write the augmented matrix for the given system of equations.
2 3 1 6
1 1 2 4
3 2 2 10
Step 2: Perform elementary row operations to simplify the matrix.
2 3 1 6
1 1 2 4
3 2 2 10
→
1 1 2 4
2 3 1 6
3 2 2 10
Step 3: Perform row operations to further simplify the matrix.
1 1 2 4
2 3 1 6
3 2 2 10
→
1 1 2 4
0 1 −3−2
0−1−4−2
Step 4: Continue to simplify the matrix using elementary row operations.
1 1 2 4
0 1 −3−2
0−1−4−2
→
1 1 2 4
0 1 −3−2
0 0 −7 0
Step 5: Write the system of equations represented by the row-echelon form
of the matrix. x+y+ 2z= 4
y−3z=−2
−7z= 0
Step 6: Solve the system of equations. From the third equation, we get
z= 0. Substitute z= 0 into the second equation to find y:
y−3(0) = −2⇒y=−2
Substitute y=−2 and z= 0 into the first equation to find x:
x−2 + 0 = 4 ⇒x= 6
Step 7: The solution to the system of equations is x= 6, y=−2, and z= 0.
Since a unique solution exists, the system is consistent. Thus, the solution to
the system is x= 6, y =−2, z = 0 .
Question 19
Question
Solve the following system of linear equations using the method of substitution:
(2x+ 3y= 11
4x−y= 5
18
Solution
Step 1: Solve the second equation for y:
4x−y= 5
y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−5) = 11
Step 3: Simplify the equation:
2x+ 12x−15 = 11
14x−15 = 11
Step 4: Solve for x:
14x= 26
x=26
14
x=13
7
Step 5: Substitute the value of xback into the equation y= 4x−5 to solve
for y:
y= 4 13
7−5
y=52
7−35
7
y=17
7
Therefore, the solution to the system of equations is x=13
7and y=17
7.
Question 20
Question
Solve the following system of linear equations:
3x+ 2y−z= 1
x−3y+ 2z=−7
2x+y+ 4z= 2
19
Solution
Step 1: Write the augmented matrix for the system of equations.
3 2 −1|1
1−3 2 | −7
2 1 4 |2
Step 2: Use row operations to get the matrix into row-echelon form.
1−0.33 0.67 | −2.33
0 1 −1|3
0 0 1 |4
Step 3: Use back-substitution to solve for the variables. From the last
equation, z= 4.
Substitute z= 4 into the second equation to solve for y:y=−1 + 4 = 3
Substitute z= 4 and y= 3 into the first equation to solve for x:x=
−2.33 + 0.33(3) −0.67(4) = −3.
Therefore, the solution to the system of equations is x=−3, y= 3, z= 4.
Question 21
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+y−z=−2
4x−2y+ 3z= 16
Solution
Step 1: Write the system of equations in augmented matrix form:
2−3 1 |7
1 1 −1| −2
4−2 3 |16
Step 2: Perform row operations to simplify the augmented matrix:
1 1 −1| −2
0−5/3 5/3|11/3
0−6 7 |24
Step 3: Perform additional row operations to simplify the matrix:
1 1 −1| −2
0 1 −1| −11/3
0 0 2 |31
20
Step 4: Use back substitution to find the values of the variables: From the
last row of the matrix, we have 2z= 31, so z= 31/2 = 15.5.
Substitute zback into the middle row of the matrix: y−15.5 = −11/3, then
y=−11/3 + 15.5 = 13/6.
Substitute yand zback into the top row of the matrix: x+13/6−15.5 = −2,
then x=−2−13/6 + 15.5 = −1/6.
Step 5: The solution to the system of equations is x=−1/6, y= 13/6,
z= 31/2.
Question 22
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+ 2y−2z=−4
3x−y+ 2z= 8
Solution
Step 1: Express the system of equations in matrix form as AX =Bwhere A
is the coefficient matrix, Xis the matrix of variables, and Bis the matrix of
constants.
A=
2−3 1
1 2 −2
3−1 2
, X =
x
y
z
, B =
7
−4
8
Step 2: To solve for X, we can find the inverse of matrix Aif it exists.
Compute the determinant of A:
det(A) =
2−3 1
1 2 −2
3−1 2
= 2(2(2) −(−2)(−1)) −(−3(2) −1(3)) + 1(1 −6) = 21
Since det(A)= 0, matrix Ais invertible. Now, find the inverse of Ausing
the formula A−1=1
det(A)adj(A) where adj(A) is the adjugate of matrix A:
adj(A) =
2 13 7
−3 11 5
−7−5 4
A−1=1
21
2 13 7
−3 11 5
−7−5 4
21
Step 3: Finally, to solve for X, multiply A−1by B:
X=A−1B=1
21
2 13 7
−3 11 5
−7−5 4
7
−4
8
=
1
−2
3
Therefore, the solution to the system of equations is x= 1, y=−2, and
z= 3.
Question 23
Question
Consider the following system of linear equations:
2x−4y+z= 5
x+ 3y−2z=−4
3x−7y+ 4z= 10
Determine if the system has a unique solution, infinitely many solutions, or no
solution. If a unique solution exists, find the solution.
Solution
Step 1: Write the augmented matrix for the system of linear equations:
2−4 1 |5
1 3 −2| −4
3−7 4 |10
Step 2: Perform elementary row operations to row-reduce the augmented
matrix:
100|3
010| −2
001|1
Step 3: Since we obtained the identity matrix on the left side of the vertical
bar, the system has a unique solution. The solution to the system of linear
equations is x= 3, y=−2, and z= 1.
Question 24
Question
Solve the system of linear equations:
2x−3y+z= 4
x−2y+ 3z= 8
3x−y−2z= 0
22
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |4
1−2 3 |8
3−1−2|0
Step 2: Use row operations to simplify the matrix. First, replace R2 with
R2−0.5R1 and R3 with R3−1.5R1:
2−3 1 |4
0−1.5 2.5|6
0 3.5−3.5| −6
Step 3: Next, replace R3 with R3+2.33R2 to get a triangular matrix:
2−3 1 |4
0−1.5 2.5|6
0 0 0 |0
Step 4: Now, back substitute to find the values of x,y, and z: From the
last row, we get 0 = 0, indicating that there are infinitely many solutions. Let
z=tbe a parameter, then:
−1.5y+ 2.5z= 6
−1.5y+ 2.5t= 6
−1.5y= 6 −2.5t
y=2.5t−6
1.5=5
3t−4
From the first row:
2x−3y+z= 4
2x−35
3t−4+t= 4
2x−5t+ 12 + t= 4
2x−4t=−8
x= 2t−4
Therefore, the solution to the system is x= 2t−4, y=5
3t−4, and z=t, where
tis a parameter.
23
Question 25
Question
Determine whether the system of linear equations below has a unique solution,
infinitely many solutions, or no solution:
3x−4y+ 2z= 7
2x+y−z=−1
x−3y+ 5z= 6
Solution
Step 1: Write the system of equations in matrix form Ax = b, where Ais the
coefficient matrix, xis the variable matrix, and bis the constant matrix.
3−4 2
2 1 −1
1−3 5
x
y
z
=
7
−1
6
Step 2: Find the determinant of matrix A. If det(A) 0, the system has a
unique solution. If det(A) = 0 and the system is consistent (i.e., bis not in the
column space of A), then the system has no solution. If det(A) = 0 and the
system is inconsistent (i.e., bis in the column space of A), then the system has
infinitely many solutions.
Step 3: Calculate the determinant of matrix Aand determine the solution
to the system of linear equations.
det(A) =
3−4 2
2 1 −1
1−3 5
= 3(1(5)−(−1)(−3))−(−4)(2(5)−(−1)(1))+2(2(−3)−1(1))
det(A) = 3(5 + 3) −(−4)(10 + 1) + 2(−6−1) = 24
Since det(A) = 24 0, the system of linear equations has a unique solution.
Question 26
Question
Solve the following system of linear equations:
2x−y+z= 4
x+ 3y−2z=−6
3x+y+ 4z= 12
24
Solution
Step 1: We can express the system of equations in matrix form AX =B, where:
A=
2−1 1
1 3 −2
3 1 4
, X =
x
y
z
,and B=
4
−6
12
Step 2: To solve for X, we can find the inverse of Aand multiply it by B:
X=A−1B
Step 3: First, we need to find the inverse of matrix A. Let’s find det(A)
first:
det(A) = (2)(3)(4)+(1)(−2)(3)+(1)(1)(−1)−(1)(3)(4)−(2)(−2)(3)−(3)(1)(1) = 26
Step 4: Now we can find A−1using the formula A−1=1
det(A)adj(A), where
adj(A) is the adjugate of matrix A.
Step 5: The adjugate of Ais found by taking the transpose of the matrix of
cofactors of A:
adj(A) =
C11 C21 C31
C12 C22 C32
C13 C23 C33
T
Step 6: Calculate the cofactors Cij for each element of A:
C11 = (−1)1+1
3−2
1 4
= (3)(4) −(−2)(1) = 14
C12 = (−1)1+2
1−2
3 4
= (1)(4) −(−2)(3) = 10
C13 = (−1)1+3
1 3
3 1
= (1)(1) −(3)(3) = −8
C21 = (−1)2+1
−1 1
1 4
= (−1)(4) −(1)(1) = −5
C22 = (−1)2+2
2 1
3 4
= (2)(4) −(1)(3) = 5
C23 = (−1)2+3
2−1
3 1
= (2)(1) −(−1)(3) = 5
C31 = (−1)3+1
−1 1
3−2
= (−1)(−2) −(1)(3) = 1
C32 = (−1)3+2
2 1
1−2
= (2)(−2) −(1)(1) = −5
C33 = (−1)3+3
2−1
1 3
= (2)(3) −(−1)(1) = 7
Step 7: Now, construct the adjugate matrix using
25
Question 27
Question
Solve the following system of linear equations using the Gauss-Jordan elimina-
tion method:
2x−3y+z= 7
4x+y−2z=−1
3x−2y+ 3z= 10
Solution
Step 1: Write the augmented matrix for the system of equations.
2−3 1 |7
4 1 −2| −1
3−2 3 |10
Step 2: Perform row operations to obtain zeros below the diagonal in the
first column.
2−3 1 |7
0 7.5−4| −15.5
0 1.5 2 | −11
Step 3: Perform row operations to obtain zeros above the diagonal in the
second column.
2 0 −3| −1
0 7.5−4| −15.5
004.5| −30
Step 4: Perform row operations to obtain a leading 1 in each column.
1 0 −1.5| −0.5
0 1 −0.53 | −2.06
0 0 1 | −6.67
Step 5: Write the system of equations corresponding to the row-echelon form
of the augmented matrix.
x−1.5z=−0.5
y−0.53z=−2.06
z=−6.67
Step 6: Solve the system of equations to find the values of x,y, and
z. From the third equation, we have z=−6.67. Substitute z=−6.67
into the first and second equations to find xand y: From the first equation,
x−1.5(−6.67) = −0.5⇒x+ 10 = −0.5⇒x=−10.5 From the second
equation, y−0.53(−6.67) = −2.06 ⇒y+ 3.54 = −2.06 ⇒y=−5.6
Therefore, the solution to the system of equations is x=−10.5, y=−5.6,
and z=−6.67.
26
Question 28
Question
Solve the following system of linear equations:
2x−3y+z= 5
3x+y−2z=−3
x+ 2y−3z= 1
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |5
3 1 −2| −3
1 2 −3|1
Step 2: Perform row operations to simplify the augmented matrix.
2−3 1 |5
3 1 −2| −3
1 2 −3|1
→
1 2 −3|1
3 1 −2| −3
2−3 1 |5
Step 3: Perform row operations to further simplify the augmented matrix.
1 2 −3|1
3 1 −2| −3
2−3 1 |5
→
1 2 −3|1
0−5 7 | −6
0−7 7 |3
Step 4: Perform row operations to continue simplifying the augmented ma-
trix.
1 2 −3|1
0−5 7 | −6
0−7 7 |3
→
1 2 −3|1
0 1 −7
5|6
5
0−7 7 |3
Step 5: Perform row operations to obtain the row-echelon form of the aug-
mented matrix.
1 2 −3|1
0 1 −7
5|6
5
0−7 7 |3
→
1 0 1
5| −7
5
0 1 −7
5|6
5
0 0 0 |9
Step 6: Write the system of equations corresponding to the row-echelon
form:
x+1
5z=−7
5
y−7
5z=6
5
0=9
Step 7: The system of equations is inconsistent since 0 = 9. Therefore, there
is no solution to the system of equations.
27
Question 29
Question
Solve the system of linear equations:
3x+ 2y−z= 2
2x−y+ 3z=−1
x+ 4y+ 4z= 5
Solution
Step 1: Write the system of equations in matrix form. Let Abe the coefficient
matrix, Xbe the unknown matrix, and Bbe the constant matrix.
A=
3 2 −1
2−1 3
1 4 4
X=
x
y
z
B=
2
−1
5
Step 2: Set up the augmented matrix [A|B].
3 2 −1 2
2−1 3 −1
1 4 4 5
Step 3: Perform row operations to row reduce the augmented matrix to
row-echelon form.
1 0 0 3
0 1 0 1
001−2
Step 4: Write down the system of equations using the row-echelon form.
x= 3
y= 1
z=−2
Step 5: So, the solution to the system of equations is x= 3, y= 1, and
z=−2.
28
Question 30
Question
Solve the following system of linear equations:
2x+ 4y−z= 7
x−2y+ 3z=−4
3x+ 2y+ 2z= 6
Solution
Step 1: Rewrite the system of equations in matrix form:
2 4 −1
1−2 3
3 2 2
x
y
z
=
7
−4
6
Step 2: Use Gaussian elimination to row reduce the augmented matrix:
2 4 −1|7
1−2 3 | −4
3 2 2 |6
Step 3: Perform row operations to obtain row-echelon form:
1−2 3 | −4
0 8 −7|15
0 8 −7|18
Step 4: Continue row operations to get reduced row-echelon form:
1−2 3 | −4
0 1 −7/8|15/8
0 0 0 |3/2
Step 5: Write the system of equations in parametric vector form:
x=−4+2s, y =15
8+7
8s, z = 0 + 0s
Step 6: Therefore, the solution to the system of equations is:
x=−4+2s, y =15
8+7
8s, z = 0
29
Question 31
Question
Solve the following system of linear equations:
3x−2y+ 4z= 7
2x+y−3z=−5
x+ 5y−2z= 9
Solution
Step 1: Let’s represent the system of equations as an augmented matrix.
3−2 4 |7
2 1 −3| −5
1 5 −2|9
Step 2: Perform row operations to simplify the matrix.
3−2 4 |7
2 1 −3| −5
1 5 −2|9
→
1 5 −2|9
2 1 −3| −5
3−2 4 |7
Step 3: Using the first row to eliminate the entries below the 1 in the first
column.
1 5 −2|9
2 1 −3| −5
3−2 4 |7
→
1 5 −2|9
0−9 1 | −23
0−17 10 | −20
Step 4: Using the second row to eliminate the entries below the -9 in the
second column.
1 5 −2|9
0−9 1 | −23
0−17 10 | −20
→
1 5 −2|9
0−9 1 | −23
0 0 3 |11
Step 5: Continue row operations to simplify the matrix.
1 5 −2|9
0−9 1 | −23
0 0 3 |11
→
1 5 −2|9
0−9 1 | −23
0 0 1 |11
3
Step 6: Perform back substitution to solve for z,y, and x.
z=11
3
−9y+ 1 = −23 =⇒y= 3
x+ 5(3) −211
3= 9 =⇒x=−3
Therefore, the solution to the system of equations is x=−3, y= 3, z=11
3.
30
Question 32
Question
Solve the following system of linear equations:
2x−3y+z= 7
3x+ 4y−2z=−10
x−5y+ 3z= 14
Solution
Step 1: Begin by writing the augmented matrix for the system of linear equa-
tions:
2−3 1 |7
3 4 −2| − 10
1−5 3 |14
Step 2: Perform row operations to simplify the matrix.
Step 3: Multiply the first row by −3
2and add it to the second row:
2−3 1 |7
017
2−7
2| − 31
2
1−5 3 |14
Step 4: Multiply the first row by −1
2and add it to the third row:
2−3 1 |7
017
2−7
2| − 31
2
0−1
2
5
2|10
Step 5: Multiply the second row by 2
17 :
2−3 1 |7
0 1 −7
17 | − 31
17
0−1
2
5
2|10
Step 6: Add 3
2times the second row to the third row:
2−3 1 |7
0 1 −7
17 | − 31
17
0 0 3 |9
Step 7: Divide the third row by 3:
2−3 1 |7
0 1 −7
17 | − 31
17
0 0 1 |3
31
Step 8: By back substitution, we find z= 3. Substitute z= 3 back into the
second row to find y:
y−7
17(3) = −31
17 ⇒y=−2
Step 9: Substitute y=−2 and z= 3 into the first row to find x:
2x−3(−2) + 3 = 7 ⇒x= 2
Therefore, the solution to the system of linear equations is x= 2, y=−2,
and z= 3.
Question 33
Question
Determine whether the system of linear equations is consistent, inconsistent, or
dependent:
2x−y= 3
4x−2y= 6
Solution
Step 1: Write the system of equations in matrix form.
2−1
4−2x
y=3
6
Step 2: Reduce the augmented matrix to row-echelon form.
2−1|3
4−2|6→2−1|3
0 0 |0
Step 3: Rewrite the augmented matrix as a system of equations.
2x−y= 3
0=0
Step 4: Determine the solution set of the system. The system has infinitely
many solutions since it represents a dependent system of linear equations. The
solution set can be expressed as:
x=t
y= 2t
where tis a parameter representing any real number.
32
Question 34
Question
Find the solution to the system of linear equations:
(2x−y= 4
3x+ 2y= 7
Solution
Step 1: To solve the system of equations, we can use the method of substitution.
Rearranging the first equation, we have:
y= 2x−4
Step 2: Substitute this expression for yinto the second equation:
3x+ 2(2x−4) = 7
Step 3: Simplify the equation by distributing the 2 on the second term:
3x+ 4x−8 = 7
Step 4: Combine like terms:
7x−8=7
Step 5: Add 8 to both sides of the equation:
7x= 15
Step 6: Divide by 7 to solve for x:
x=15
7
Step 7: Substitute the value of xback into the equation y= 2x−4 to find
y:
y= 2 15
7−4
Step 8: Simplify the expression to find y:
y=30
7−28
7=2
7
Step 9: Therefore, the solution to the system of equations is (x, y) = 15
7,2
7.
33
Question 35
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
5x−2y+ 4z= 7
3x+y−2z= 1
Solution
Step 1: Rewrite the system of equations in matrix form:
2 3 −1
5−2 4
3 1 −2
x
y
z
=
4
7
1
Step 2: Use row operations to simplify the augmented matrix:
2 3 −1|4
5−2 4 |7
3 1 −2|1
Step 3: Perform row operations to obtain the following row-echelon form:
100|3
010|2
001| −2
Step 4: Therefore, the solution to the system of equations is x= 3, y= 2,
and z=−2.
34
Step 3: Solve for zusing back substitution.
−29
7z=−2 =⇒z=14
29
Step 4: Substitute z=14
29 back into the second row to solve for y.
−7y+ 4 14
29= 3 =⇒y=−35
29
Step 5: Substitute z=14
29 and y=−35
29 back into the first row to solve for
x.
2x+ 3 −35
29−14
29 = 1 =⇒x=49
29
Therefore, the solution to the system of equations is:
x=49
29, y =−35
29, z =14
29
Question 2
Question
Solve the following system of linear equations:
2x−3y+z= 7
4x+y−z= 1
3x−2y+ 2z= 10
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |7
4 1 −1|1
3−2 2 |10
Step 2: Perform row operations to simplify the matrix. Start by multiplying
the first row by 2 and subtracting 4 times the first row from the second row,
and subtracting 3 times the first row from the third row to get:
2−3 1 |7
0 13 −5| −27
0 7 −1| −11
Step 3: Next, divide the second row by 13 to simplify the matrix further:
2−3 1 |7
0 1 −5
13 | −27
13
0 7 −1| −11
2
Step 4: Perform row operations to eliminate the 7 in the third row. Subtract
7 times the second row from the third row to get:
2−3 1 |7
0 1 −5
13 | −27
13
0 0 32
13 |10
Step 5: Divide the third row by 32
13 to find the value of z:
2−3 1 |7
0 1 −5
13 | −27
13
0 0 1 |130
32
Step 6: Substitute z=130
32 back into the second row to find the value of y:
1y−5
13z=−27
13y−5
13 ×130
32 =−27
13y−25
16 =−27
13y=−27
13 +25
16 =−79
26
Step 7: Finally, substitute z=130
32 and y=−79
26 back into the first row to
find the value of x:
2x−3y+z= 72x−3×−79
26+130
32 = 72x+237
26 +130
32 = 72x= 7−237
26 −130
32 2x=182
13 x=91
13
Therefore, the solution to the system of linear equations is x=91
13 ,y=−79
26 ,
and z=130
32 .
Question 3
Question
Solve the following system of equations:
2x−3y+z= 4
4x+y−2z=−3
x−2y+ 3z= 5
Solution
Step 1: Let’s write the system of equations in matrix form:
2−3 1
4 1 −2
1−2 3
x
y
z
=
4
−3
5
Step 2: We will perform row operations to solve this system of equations.
First, we will subtract the first row multiplied by 2 from the second row, and
subtract the first row multiplied by 1
2from the third row:
2−3 1
0 7 −4
0−5
2
5
2
x
y
z
=
4
−11
7
2
3
Step 3: Next, we will divide the second row by 7 and the third row by −5
2:
2−3 1
0 1 −4
7
0 1 −1
x
y
z
=
4
−11
7
−7
10
Step 4: Now, we’ll subtract the second row from the third row:
2−3 1
0 1 −4
7
0 0 −3
10
x
y
z
=
4
−11
7
1
10
Step 5: Lastly, we can back substitute to find the values of x,y, and z. We
get z=−1
3,y=−4
7, and x=11
3.
Therefore, the solution to the system of equations is x=11
3,y=−4
7, and
z=−1
3.
Question 4
Question
Solve the following system of linear equations:
2x−y+z= 3
x+ 3y−2z=−4
3x−2y+ 2z= 8
Solution
Step 1: We start by writing the system of linear equations in matrix form
Ax=b, where Ais the coefficient matrix, xis the column matrix of variables,
and bis the column matrix of constants.
2−1 1
1 3 −2
3−2 2
x
y
z
=
3
−4
8
Step 2: Next, we perform row operations to get the matrix in row-echelon
form:
2−1 1 |3
1 3 −2| −4
3−2 2 |8
Row 2 - 1
2Row 1
Row 3 - 3
2Row 1
2−1 1 |3
07
2−5
2| −11
2
01
2
1
2|11
2
4
Step 3: Continue row operations to simplify the matrix further.
Row 2 ×2
7
Row 3 - 1
2Row 2
2−1 1 |3
0 1 −1| −11
7
0 0 2 |30
7
Step 4: Finally, we convert the matrix back into equations and solve for x,
y, and z. From the last row, we have 2z=30
7, so z=15
7.
Substitute z=15
7back into the second row: y−15
7=−11
7, which gives
y= 2.
Now substitute y= 2 and z=15
7back into the first row: 2x−2 + 15
7= 3,
which yields x= 1.
Therefore, the solution to the system of linear equations is x= 1, y= 2, and
z=15
7.
Question 5
Question
Find the values of aand bsuch that the following system of linear equations
has infinitely many solutions:
2x−3y= 4
ax +by = 2
Solution
Step 1: To have infinitely many solutions, the two equations must be linearly
dependent; in other words, one equation must be a scalar multiple of the other.
This means that the ratios of the coefficients of the variables in the two equations
must be equal. Therefore, we need to find aand bsuch that:
2
a=−3
b=4
2
Step 2: From the first ratio, we have 2
a=4
2, which simplifies to 2a= 4.
Solving for a, we get a= 2.
Step 3: Now substitute a= 2 into the second ratio: −3
b=4
2, which simplifies
to −3b= 8. Solving for b, we get b=−8
3.
Step 4: Therefore, the values of aand bsuch that the system of linear
equations has infinitely many solutions are a= 2 and b=−8
3.
5
Question 6
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+y−z= 2
3x−4y+ 2z= 6
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 |4
1 1 −1|2
3−4 2 |6
Step 2: Perform row operations to simplify the matrix.
1 1 −1|2
0−5 3 |0
0−7 5 |0
Step 3: Perform further row operations to obtain the reduced row-echelon
form.
1 1 −1|2
0 1 −3
5|0
0 0 1 |0
Step 4: Back-substitute to find the values of z,y, and x. From the last
row, z= 0. Substituting z= 0 back into the second row yields:
y−3
5= 0 ⇒y=3
5
Substituting y=3
5and z= 0 into the first row gives:
x+3
5−1=2⇒x=7
5
Step 5: Therefore, the solution to the system of equations is:
x=7
5, y =3
5, z = 0
6
Question 7
Question
Solve the following system of linear equations using the method of your choice:
3x−2y+z= 7
2x+ 4y+z= 4
x−3y−z=−5
Solution
Step 1: Rewrite the system of equations in matrix form.
3−2 1
2 4 1
1−3−1
x
y
z
=
7
4
−5
Step 2: Perform row operations to simplify the matrix equation.
3−2 1 |7
2 4 1 |4
1−3−1| −5
Step 3: Perform row operations to get the matrix in row-echelon form.
1−3−1| −5
0 10 4 |19
0 1 4 |12
Step 4: Continue to row-reduce the matrix to get it in reduced row-echelon
form.
1 0 −7| −23
014/10 |19/10
0 0 1 |11
Step 5: Convert the reduced matrix back into a system of equations.
x=−23
y=19
10
z= 11
Therefore, the solution to the system of equations is x=−23, y=19
10 , and
z= 11.
7
Question 8
Question
Solve the following system of linear equations:
3x−2y+ 4z= 7
x+ 2y−z=−1
2x−3y+ 5z= 10
Solution
Step 1: To solve the system of equations, we will use the method of substitution.
First, we will solve the second equation for x:
x=−1 + z−2y
Step 2: Next, we will substitute xinto the first and third equations:
3(−1 + z−2y)−2y+ 4z= 7
2(−1 + z−2y)−3y+ 5z= 10
Step 3: Simplifying the equations, we have:
−3+3z−6y−2y+ 4z= 7
−2+2z−4y−3y+ 5z= 10
Step 4: Simplifying further, we get:
7y+ 7z= 10
7y+ 7z= 12
Step 5: It is clear that the system of equations is inconsistent, as the last
two equations are contradictory. Therefore, there is no solution to the system
of equations.
Question 9
Question
Solve the following system of linear equations using matrix methods:
2x−y+ 3z= 4
x+ 2y+z= 1
3x−2y+ 2z= 5
8
Solution
Step 1: Write the system of equations in matrix form as AX =B, where
A=
2−1 3
121
3−2 2
, X =
x
y
z
, B =
4
1
5
Step 2: Find the inverse of matrix Aby calculating |A|and A−1.
|A|= 2(2)(2)+1(−2)(3)+3(1)(−1)−3(2)(3)−1(−2)(2)−2(1)(2) = 4−6−3−18+4−4 = −23
A−1=1
|A|
213
2 2 −2
−1 1 −1
=
−2
23 −1
23 −3
23
−2
23 −2
23
2
23
1
23 −1
23
1
23
Step 3: Calculate X=A−1B
X=
−2
23 −1
23 −3
23
−2
23 −2
23
2
23
1
23 −1
23
1
23
4
1
5
=
−4
23 −1
23 −15
23
−8
23 −2
23 +10
23
4
23 +1
23 +5
23
=
−20
23
0
10
23
Therefore, the solution to the system of equations is x=−20
23 ,y= 0, and
z=10
23 .
Question 10
Question
Solve the following system of linear equations:
2x+y+z= 4
x−y+ 3z= 5
3x+ 2y−z= 1
Solution
Step 1: Rewrite the system of equations in matrix form.
2 1 1
1−1 3
3 2 −1
x
y
z
=
4
5
1
Step 2: Find the determinant of the coefficient matrix to determine if the
system has a unique solution.
det
2 1 1
1−1 3
3 2 −1
=−1−(−6) = 5 = 0
9
Since the determinant is nonzero, the system has a unique solution.
Step 3: Use Cramer’s Rule to find the solution.
x=
det
4 1 1
5−1 3
1 2 −1
det
2 1 1
1−1 3
3 2 −1
=31
5
y=
det
2 4 1
1 5 3
3 1 −1
det
2 1 1
1−1 3
3 2 −1
=17
5
z=
det
214
1−1 5
321
det
2 1 1
1−1 3
3 2 −1
=1
5
Therefore, the solution to the system of linear equations is x=31
5,y=17
5,
and z=1
5.
Question 11
Question
Solve the following system of linear equations using any method of your choice:
3x−2y+z= 4
x+ 3y−2z= 6
2x−y+ 3z= 8
Solution
Step 1: We can solve this system of linear equations using the method of sub-
stitution or elimination. Let’s use the elimination method.
Step 2: Start by eliminating the variable xfrom the system. Multiply the
first equation by 1, the second equation by 3, and leave the third equation as is:
3x−2y+z= 4
3x+ 9y−6z= 18
2x−y+ 3z= 8
10
Step 3: Subtract the first equation from the second equation to eliminate x:
11y−7z= 14
Step 4: Now, eliminate xfrom the first and third equations. Multiply the
first equation by 2, the third equation by 3, and leave the second equation as is:
6x−4y+ 2z= 8
6x−3y+ 9z= 24
2x−y+ 3z= 8
Step 5: Subtract the third equation from the first equation to eliminate x:
3y−7z= 0
Step 6: We now have a system of two equations with two variables:
(11y−7z= 14
3y−7z= 0
Step 7: Solve this system to find yand z. After finding the values of yand
z, substitute them back into any of the original equations to solve for x.
Step 8: Finally, write the solution as an ordered triple (x, y, z).
Question 12
Question
Solve the following system of linear equations using the Gaussian elimination
method:
2x+5y−z= 9
x+3y+2z= 17
3x+7y+4z= 25
Solution
Step 1: Write the augmented matrix for the system of equations.
2 5 −1|9
1 3 2 |17
3 7 4 |25
Step 2: Perform row operations to simplify the matrix.
2 5 −1|9
1 3 2 |17
3 7 4 |25
→
1 3 2 |17
2 5 −1|9
3 7 4 |25
11
Step 3: Perform row operations to simplify the matrix further.
1 3 2 |17
2 5 −1|9
3 7 4 |25
→
1 3 2 |17
0−1−5| −25
0−2−2| −34
Step 4: Perform row operations to get the matrix in row-echelon form.
1 3 2 |17
0−1−5| −25
0−2−2| −34
→
1 3 2 |17
0 1 5 |25
0−2−2| −34
Step 5: Perform row operations to obtain the reduced row-echelon form.
1 3 2 |17
0 1 5 |25
0−2−2| −34
→
1 0 −13 | −34
0 1 5 |25
0 0 8 |16
Step 6: Solve for the variables. From the row-echelon form, we have:
x=−34, y = 25, z = 2
Therefore, the solution to the system of equations is x=−34, y= 25, and
z= 2.
Question 13
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
x−2y+ 2z=−3
3x+ 2y−3z= 5
Solution
Step 1: We will solve the system of equations using the method of substitution.
Step 2: From the first equation, we can express zin terms of xand y:
z= 2x+ 3y−4
Step 3: Substitute the expression for zinto the second and third equations:
x−2y+ 2(2x+ 3y−4) = −3
3x+ 2y−3(2x+ 3y−4) = 5
12
Step 4: Simplify the equations:
x−2y+ 4x+ 6y−8 = −3
3x+ 2y−6x−9y+ 12 = 5
Step 5: Combine like terms:
5x+ 4y= 5
−3x−7y=−7
Step 6: Solve this system of equations. Multiplying the first equation by 3 and
the second equation by 5 to eliminate ygives:
15x+ 12y= 15
−15x−35y=−35
Step 7: Adding the two equations results in:
−23y=−20
Step 8: Solve for y:
y=20
23
Step 9: Substitute yback into the equation 5x+ 4y= 5 to solve for x:
5x+ 4 20
23= 5
5x+80
23 = 5
5x=115
23
x=23
23 = 1
Step 10: Substitute xand yback into the equation z= 2x+ 3y−4 to solve for
z:
z= 2(1) + 3 20
23−4 = 2 + 60
23 −92
23 =−30
23
Step 11: Therefore, the solution to the system of equations is x= 1, y=20
23 ,
and z=−30
23 .
Question 14
Question
Solve the following system of linear equations:
2x−3y+z= 1
x+ 4y−2z=−2
3x−2y+ 5z= 10
13
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−3 1 |1
1 4 −2| −2
3−2 5 |10
Step 2: Perform row operations to row-echelon form:
100|3
010| −1
001| −2
Step 3: The solution to the system is x= 3, y=−1, and z=−2.
Question 15
Question
Solve the following system of linear equations:
2x−y+ 3z= 5
x+y+ 2z= 4
3x+ 5y−z= 2
Solution
Step 1: Write the system of equations in matrix form. Let’s denote the coeffi-
cient matrix as A, the variable matrix as X, and the constant matrix as B. The
system AX =Bcan be represented as:
2−1 3
1 1 2
3 5 −1
x
y
z
=
5
4
2
Step 2: Use Gaussian elimination to row-reduce the augmented matrix [A|B].
2−1 3 |5
1 1 2 |4
3 5 −1|2
Step 3: Perform row operations to row-reduce the augmented matrix:
1 1 2 |4
0 3 −1|1
0 2 −7| −10
14
Step 4: Continue row operations:
1 1 2 |4
0 1 −1
3|1
3
0 0 −19
3| −11
3
Step 5: Write the system of equations back with row-reduced form:
x+y+ 2z= 4
y−1
3z=1
3
−19
3z=−11
3
Step 6: Solve for z:z=−11
19
Step 7: Substitute zback into the second equation to solve for y:y−1
3·
−11
19 =1
3
y+11
57 =19
57
y=8
19
Step 8: Substitute yand zback into the first equation to solve for x:x+
8
19 + 2 ·−11
19 = 4
x=35
19
Therefore, the solution to the system of equations is x=35
19 ,y=8
19 ,z=−11
19 .
Question 16
Question
Solve the following system of linear equations using matrix methods:
2x−3y+z= 5
x+y+ 2z= 6
3x−2y−3z=−7
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
1 1 2
3−2−3
x
y
z
=
5
6
−7
Step 2: Set up the augmented matrix and perform row operations.
[ccc|c]2 −3 1 5
1 1 2 6
3−2−3−7
15
Step 3: Perform row operations to get the matrix in row-echelon form.
[ccc|c]1 1 2 6
0−5−3−7
0 0 0 0
Step 4: Solve for the variables starting from the bottom row.
−5y−3z=−7 =⇒y= 1 −3
5z
Step 5: Substitute yinto the next equation and solve for x.
x+ 1 −3
5z+ 2z= 6 =⇒x= 2 + 8
5z
Step 6: Substitute xand yback into the first equation to find z.
2(2 + 8
5z)−3(1 −3
5z) + z= 5 =⇒z= 5
Step 7: Finally, substitute zback into the expressions for xand yto get
the solution.
x= 2 + 8
5(5) = 10, y = 1 −3
5(5) = −2
Therefore, the solution to the system of equations is x= 10, y=−2, and
z= 5.
Question 17
Question
Solve the following system of linear equations:
3x−2y+z= 5
x+ 4y−2z=−10
2x−y+ 3z= 4
Solution
Step 1: Write the system of equations in matrix form.
3−2 1
1 4 −2
2−1 3
x
y
z
=
5
−10
4
Step 2: Find the inverse of the matrix on the left side of the equation.
Let A=
3−2 1
1 4 −2
2−1 3
16
To find A−1, we will row reduce the augmented matrix [A — I].
3−2 1 |100
1 4 −2|010
2−1 3 |001
Step 3: Perform row operations to convert the left side to the identity matrix.
100|5
17 −3
17
1
17
010| − 2
17
1
17
1
17
001| − 6
17
2
17
1
17
Therefore, the inverse of matrix Ais:
A−1=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
Step 4: Calculate the solution vector by multiplying A−1with the column
vector on the right side of the equation.
x
y
z
=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
5
−10
4
Step 5: Perform matrix multiplication to find the solution vector.
x
y
z
=
5
17 −3
17
1
17
−2
17
1
17
1
17
−6
17
2
17
1
17
5
−10
4
=
7
17
−12
17
1
17
Therefore, the solution to the system of equations is x=7
17 ,y=−12
17 , and
z=1
17 .
Question 18
Question
Consider the following system of linear equations:
2x+ 3y+z= 6
x+y+ 2z= 4
3x+ 2y+ 2z= 10
Determine if the system of equations is consistent. If it is consistent, find
the solution to the system. If it is inconsistent, explain why.
17
Solution
Step 1: Write the augmented matrix for the given system of equations.
2 3 1 6
1 1 2 4
3 2 2 10
Step 2: Perform elementary row operations to simplify the matrix.
2 3 1 6
1 1 2 4
3 2 2 10
→
1 1 2 4
2 3 1 6
3 2 2 10
Step 3: Perform row operations to further simplify the matrix.
1 1 2 4
2 3 1 6
3 2 2 10
→
1 1 2 4
0 1 −3−2
0−1−4−2
Step 4: Continue to simplify the matrix using elementary row operations.
1 1 2 4
0 1 −3−2
0−1−4−2
→
1 1 2 4
0 1 −3−2
0 0 −7 0
Step 5: Write the system of equations represented by the row-echelon form
of the matrix. x+y+ 2z= 4
y−3z=−2
−7z= 0
Step 6: Solve the system of equations. From the third equation, we get
z= 0. Substitute z= 0 into the second equation to find y:
y−3(0) = −2⇒y=−2
Substitute y=−2 and z= 0 into the first equation to find x:
x−2 + 0 = 4 ⇒x= 6
Step 7: The solution to the system of equations is x= 6, y=−2, and z= 0.
Since a unique solution exists, the system is consistent. Thus, the solution to
the system is x= 6, y =−2, z = 0 .
Question 19
Question
Solve the following system of linear equations using the method of substitution:
(2x+ 3y= 11
4x−y= 5
18
Solution
Step 1: Solve the second equation for y:
4x−y= 5
y= 4x−5
Step 2: Substitute the expression for yinto the first equation:
2x+ 3(4x−5) = 11
Step 3: Simplify the equation:
2x+ 12x−15 = 11
14x−15 = 11
Step 4: Solve for x:
14x= 26
x=26
14
x=13
7
Step 5: Substitute the value of xback into the equation y= 4x−5 to solve
for y:
y= 4 13
7−5
y=52
7−35
7
y=17
7
Therefore, the solution to the system of equations is x=13
7and y=17
7.
Question 20
Question
Solve the following system of linear equations:
3x+ 2y−z= 1
x−3y+ 2z=−7
2x+y+ 4z= 2
19
Solution
Step 1: Write the augmented matrix for the system of equations.
3 2 −1|1
1−3 2 | −7
2 1 4 |2
Step 2: Use row operations to get the matrix into row-echelon form.
1−0.33 0.67 | −2.33
0 1 −1|3
0 0 1 |4
Step 3: Use back-substitution to solve for the variables. From the last
equation, z= 4.
Substitute z= 4 into the second equation to solve for y:y=−1 + 4 = 3
Substitute z= 4 and y= 3 into the first equation to solve for x:x=
−2.33 + 0.33(3) −0.67(4) = −3.
Therefore, the solution to the system of equations is x=−3, y= 3, z= 4.
Question 21
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+y−z=−2
4x−2y+ 3z= 16
Solution
Step 1: Write the system of equations in augmented matrix form:
2−3 1 |7
1 1 −1| −2
4−2 3 |16
Step 2: Perform row operations to simplify the augmented matrix:
1 1 −1| −2
0−5/3 5/3|11/3
0−6 7 |24
Step 3: Perform additional row operations to simplify the matrix:
1 1 −1| −2
0 1 −1| −11/3
0 0 2 |31
20
Step 4: Use back substitution to find the values of the variables: From the
last row of the matrix, we have 2z= 31, so z= 31/2 = 15.5.
Substitute zback into the middle row of the matrix: y−15.5 = −11/3, then
y=−11/3 + 15.5 = 13/6.
Substitute yand zback into the top row of the matrix: x+13/6−15.5 = −2,
then x=−2−13/6 + 15.5 = −1/6.
Step 5: The solution to the system of equations is x=−1/6, y= 13/6,
z= 31/2.
Question 22
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+ 2y−2z=−4
3x−y+ 2z= 8
Solution
Step 1: Express the system of equations in matrix form as AX =Bwhere A
is the coefficient matrix, Xis the matrix of variables, and Bis the matrix of
constants.
A=
2−3 1
1 2 −2
3−1 2
, X =
x
y
z
, B =
7
−4
8
Step 2: To solve for X, we can find the inverse of matrix Aif it exists.
Compute the determinant of A:
det(A) =
2−3 1
1 2 −2
3−1 2
= 2(2(2) −(−2)(−1)) −(−3(2) −1(3)) + 1(1 −6) = 21
Since det(A)= 0, matrix Ais invertible. Now, find the inverse of Ausing
the formula A−1=1
det(A)adj(A) where adj(A) is the adjugate of matrix A:
adj(A) =
2 13 7
−3 11 5
−7−5 4
A−1=1
21
2 13 7
−3 11 5
−7−5 4
21
Step 3: Finally, to solve for X, multiply A−1by B:
X=A−1B=1
21
2 13 7
−3 11 5
−7−5 4
7
−4
8
=
1
−2
3
Therefore, the solution to the system of equations is x= 1, y=−2, and
z= 3.
Question 23
Question
Consider the following system of linear equations:
2x−4y+z= 5
x+ 3y−2z=−4
3x−7y+ 4z= 10
Determine if the system has a unique solution, infinitely many solutions, or no
solution. If a unique solution exists, find the solution.
Solution
Step 1: Write the augmented matrix for the system of linear equations:
2−4 1 |5
1 3 −2| −4
3−7 4 |10
Step 2: Perform elementary row operations to row-reduce the augmented
matrix:
100|3
010| −2
001|1
Step 3: Since we obtained the identity matrix on the left side of the vertical
bar, the system has a unique solution. The solution to the system of linear
equations is x= 3, y=−2, and z= 1.
Question 24
Question
Solve the system of linear equations:
2x−3y+z= 4
x−2y+ 3z= 8
3x−y−2z= 0
22
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |4
1−2 3 |8
3−1−2|0
Step 2: Use row operations to simplify the matrix. First, replace R2 with
R2−0.5R1 and R3 with R3−1.5R1:
2−3 1 |4
0−1.5 2.5|6
0 3.5−3.5| −6
Step 3: Next, replace R3 with R3+2.33R2 to get a triangular matrix:
2−3 1 |4
0−1.5 2.5|6
0 0 0 |0
Step 4: Now, back substitute to find the values of x,y, and z: From the
last row, we get 0 = 0, indicating that there are infinitely many solutions. Let
z=tbe a parameter, then:
−1.5y+ 2.5z= 6
−1.5y+ 2.5t= 6
−1.5y= 6 −2.5t
y=2.5t−6
1.5=5
3t−4
From the first row:
2x−3y+z= 4
2x−35
3t−4+t= 4
2x−5t+ 12 + t= 4
2x−4t=−8
x= 2t−4
Therefore, the solution to the system is x= 2t−4, y=5
3t−4, and z=t, where
tis a parameter.
23
Question 25
Question
Determine whether the system of linear equations below has a unique solution,
infinitely many solutions, or no solution:
3x−4y+ 2z= 7
2x+y−z=−1
x−3y+ 5z= 6
Solution
Step 1: Write the system of equations in matrix form Ax = b, where Ais the
coefficient matrix, xis the variable matrix, and bis the constant matrix.
3−4 2
2 1 −1
1−3 5
x
y
z
=
7
−1
6
Step 2: Find the determinant of matrix A. If det(A) 0, the system has a
unique solution. If det(A) = 0 and the system is consistent (i.e., bis not in the
column space of A), then the system has no solution. If det(A) = 0 and the
system is inconsistent (i.e., bis in the column space of A), then the system has
infinitely many solutions.
Step 3: Calculate the determinant of matrix Aand determine the solution
to the system of linear equations.
det(A) =
3−4 2
2 1 −1
1−3 5
= 3(1(5)−(−1)(−3))−(−4)(2(5)−(−1)(1))+2(2(−3)−1(1))
det(A) = 3(5 + 3) −(−4)(10 + 1) + 2(−6−1) = 24
Since det(A) = 24 0, the system of linear equations has a unique solution.
Question 26
Question
Solve the following system of linear equations:
2x−y+z= 4
x+ 3y−2z=−6
3x+y+ 4z= 12
24
Solution
Step 1: We can express the system of equations in matrix form AX =B, where:
A=
2−1 1
1 3 −2
3 1 4
, X =
x
y
z
,and B=
4
−6
12
Step 2: To solve for X, we can find the inverse of Aand multiply it by B:
X=A−1B
Step 3: First, we need to find the inverse of matrix A. Let’s find det(A)
first:
det(A) = (2)(3)(4)+(1)(−2)(3)+(1)(1)(−1)−(1)(3)(4)−(2)(−2)(3)−(3)(1)(1) = 26
Step 4: Now we can find A−1using the formula A−1=1
det(A)adj(A), where
adj(A) is the adjugate of matrix A.
Step 5: The adjugate of Ais found by taking the transpose of the matrix of
cofactors of A:
adj(A) =
C11 C21 C31
C12 C22 C32
C13 C23 C33
T
Step 6: Calculate the cofactors Cij for each element of A:
C11 = (−1)1+1
3−2
1 4
= (3)(4) −(−2)(1) = 14
C12 = (−1)1+2
1−2
3 4
= (1)(4) −(−2)(3) = 10
C13 = (−1)1+3
1 3
3 1
= (1)(1) −(3)(3) = −8
C21 = (−1)2+1
−1 1
1 4
= (−1)(4) −(1)(1) = −5
C22 = (−1)2+2
2 1
3 4
= (2)(4) −(1)(3) = 5
C23 = (−1)2+3
2−1
3 1
= (2)(1) −(−1)(3) = 5
C31 = (−1)3+1
−1 1
3−2
= (−1)(−2) −(1)(3) = 1
C32 = (−1)3+2
2 1
1−2
= (2)(−2) −(1)(1) = −5
C33 = (−1)3+3
2−1
1 3
= (2)(3) −(−1)(1) = 7
Step 7: Now, construct the adjugate matrix using
25
Question 27
Question
Solve the following system of linear equations using the Gauss-Jordan elimina-
tion method:
2x−3y+z= 7
4x+y−2z=−1
3x−2y+ 3z= 10
Solution
Step 1: Write the augmented matrix for the system of equations.
2−3 1 |7
4 1 −2| −1
3−2 3 |10
Step 2: Perform row operations to obtain zeros below the diagonal in the
first column.
2−3 1 |7
0 7.5−4| −15.5
0 1.5 2 | −11
Step 3: Perform row operations to obtain zeros above the diagonal in the
second column.
2 0 −3| −1
0 7.5−4| −15.5
004.5| −30
Step 4: Perform row operations to obtain a leading 1 in each column.
1 0 −1.5| −0.5
0 1 −0.53 | −2.06
0 0 1 | −6.67
Step 5: Write the system of equations corresponding to the row-echelon form
of the augmented matrix.
x−1.5z=−0.5
y−0.53z=−2.06
z=−6.67
Step 6: Solve the system of equations to find the values of x,y, and
z. From the third equation, we have z=−6.67. Substitute z=−6.67
into the first and second equations to find xand y: From the first equation,
x−1.5(−6.67) = −0.5⇒x+ 10 = −0.5⇒x=−10.5 From the second
equation, y−0.53(−6.67) = −2.06 ⇒y+ 3.54 = −2.06 ⇒y=−5.6
Therefore, the solution to the system of equations is x=−10.5, y=−5.6,
and z=−6.67.
26
Question 28
Question
Solve the following system of linear equations:
2x−3y+z= 5
3x+y−2z=−3
x+ 2y−3z= 1
Solution
Step 1: Write the system of equations as an augmented matrix:
2−3 1 |5
3 1 −2| −3
1 2 −3|1
Step 2: Perform row operations to simplify the augmented matrix.
2−3 1 |5
3 1 −2| −3
1 2 −3|1
→
1 2 −3|1
3 1 −2| −3
2−3 1 |5
Step 3: Perform row operations to further simplify the augmented matrix.
1 2 −3|1
3 1 −2| −3
2−3 1 |5
→
1 2 −3|1
0−5 7 | −6
0−7 7 |3
Step 4: Perform row operations to continue simplifying the augmented ma-
trix.
1 2 −3|1
0−5 7 | −6
0−7 7 |3
→
1 2 −3|1
0 1 −7
5|6
5
0−7 7 |3
Step 5: Perform row operations to obtain the row-echelon form of the aug-
mented matrix.
1 2 −3|1
0 1 −7
5|6
5
0−7 7 |3
→
1 0 1
5| −7
5
0 1 −7
5|6
5
0 0 0 |9
Step 6: Write the system of equations corresponding to the row-echelon
form:
x+1
5z=−7
5
y−7
5z=6
5
0=9
Step 7: The system of equations is inconsistent since 0 = 9. Therefore, there
is no solution to the system of equations.
27
Question 29
Question
Solve the system of linear equations:
3x+ 2y−z= 2
2x−y+ 3z=−1
x+ 4y+ 4z= 5
Solution
Step 1: Write the system of equations in matrix form. Let Abe the coefficient
matrix, Xbe the unknown matrix, and Bbe the constant matrix.
A=
3 2 −1
2−1 3
1 4 4
X=
x
y
z
B=
2
−1
5
Step 2: Set up the augmented matrix [A|B].
3 2 −1 2
2−1 3 −1
1 4 4 5
Step 3: Perform row operations to row reduce the augmented matrix to
row-echelon form.
1 0 0 3
0 1 0 1
001−2
Step 4: Write down the system of equations using the row-echelon form.
x= 3
y= 1
z=−2
Step 5: So, the solution to the system of equations is x= 3, y= 1, and
z=−2.
28
Question 30
Question
Solve the following system of linear equations:
2x+ 4y−z= 7
x−2y+ 3z=−4
3x+ 2y+ 2z= 6
Solution
Step 1: Rewrite the system of equations in matrix form:
2 4 −1
1−2 3
3 2 2
x
y
z
=
7
−4
6
Step 2: Use Gaussian elimination to row reduce the augmented matrix:
2 4 −1|7
1−2 3 | −4
3 2 2 |6
Step 3: Perform row operations to obtain row-echelon form:
1−2 3 | −4
0 8 −7|15
0 8 −7|18
Step 4: Continue row operations to get reduced row-echelon form:
1−2 3 | −4
0 1 −7/8|15/8
0 0 0 |3/2
Step 5: Write the system of equations in parametric vector form:
x=−4+2s, y =15
8+7
8s, z = 0 + 0s
Step 6: Therefore, the solution to the system of equations is:
x=−4+2s, y =15
8+7
8s, z = 0
29
Question 31
Question
Solve the following system of linear equations:
3x−2y+ 4z= 7
2x+y−3z=−5
x+ 5y−2z= 9
Solution
Step 1: Let’s represent the system of equations as an augmented matrix.
3−2 4 |7
2 1 −3| −5
1 5 −2|9
Step 2: Perform row operations to simplify the matrix.
3−2 4 |7
2 1 −3| −5
1 5 −2|9
→
1 5 −2|9
2 1 −3| −5
3−2 4 |7
Step 3: Using the first row to eliminate the entries below the 1 in the first
column.
1 5 −2|9
2 1 −3| −5
3−2 4 |7
→
1 5 −2|9
0−9 1 | −23
0−17 10 | −20
Step 4: Using the second row to eliminate the entries below the -9 in the
second column.
1 5 −2|9
0−9 1 | −23
0−17 10 | −20
→
1 5 −2|9
0−9 1 | −23
0 0 3 |11
Step 5: Continue row operations to simplify the matrix.
1 5 −2|9
0−9 1 | −23
0 0 3 |11
→
1 5 −2|9
0−9 1 | −23
0 0 1 |11
3
Step 6: Perform back substitution to solve for z,y, and x.
z=11
3
−9y+ 1 = −23 =⇒y= 3
x+ 5(3) −211
3= 9 =⇒x=−3
Therefore, the solution to the system of equations is x=−3, y= 3, z=11
3.
30
Question 32
Question
Solve the following system of linear equations:
2x−3y+z= 7
3x+ 4y−2z=−10
x−5y+ 3z= 14
Solution
Step 1: Begin by writing the augmented matrix for the system of linear equa-
tions:
2−3 1 |7
3 4 −2| − 10
1−5 3 |14
Step 2: Perform row operations to simplify the matrix.
Step 3: Multiply the first row by −3
2and add it to the second row:
2−3 1 |7
017
2−7
2| − 31
2
1−5 3 |14
Step 4: Multiply the first row by −1
2and add it to the third row:
2−3 1 |7
017
2−7
2| − 31
2
0−1
2
5
2|10
Step 5: Multiply the second row by 2
17 :
2−3 1 |7
0 1 −7
17 | − 31
17
0−1
2
5
2|10
Step 6: Add 3
2times the second row to the third row:
2−3 1 |7
0 1 −7
17 | − 31
17
0 0 3 |9
Step 7: Divide the third row by 3:
2−3 1 |7
0 1 −7
17 | − 31
17
0 0 1 |3
31
Step 8: By back substitution, we find z= 3. Substitute z= 3 back into the
second row to find y:
y−7
17(3) = −31
17 ⇒y=−2
Step 9: Substitute y=−2 and z= 3 into the first row to find x:
2x−3(−2) + 3 = 7 ⇒x= 2
Therefore, the solution to the system of linear equations is x= 2, y=−2,
and z= 3.
Question 33
Question
Determine whether the system of linear equations is consistent, inconsistent, or
dependent:
2x−y= 3
4x−2y= 6
Solution
Step 1: Write the system of equations in matrix form.
2−1
4−2x
y=3
6
Step 2: Reduce the augmented matrix to row-echelon form.
2−1|3
4−2|6→2−1|3
0 0 |0
Step 3: Rewrite the augmented matrix as a system of equations.
2x−y= 3
0=0
Step 4: Determine the solution set of the system. The system has infinitely
many solutions since it represents a dependent system of linear equations. The
solution set can be expressed as:
x=t
y= 2t
where tis a parameter representing any real number.
32
Question 34
Question
Find the solution to the system of linear equations:
(2x−y= 4
3x+ 2y= 7
Solution
Step 1: To solve the system of equations, we can use the method of substitution.
Rearranging the first equation, we have:
y= 2x−4
Step 2: Substitute this expression for yinto the second equation:
3x+ 2(2x−4) = 7
Step 3: Simplify the equation by distributing the 2 on the second term:
3x+ 4x−8 = 7
Step 4: Combine like terms:
7x−8=7
Step 5: Add 8 to both sides of the equation:
7x= 15
Step 6: Divide by 7 to solve for x:
x=15
7
Step 7: Substitute the value of xback into the equation y= 2x−4 to find
y:
y= 2 15
7−4
Step 8: Simplify the expression to find y:
y=30
7−28
7=2
7
Step 9: Therefore, the solution to the system of equations is (x, y) = 15
7,2
7.
33
Question 35
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
5x−2y+ 4z= 7
3x+y−2z= 1
Solution
Step 1: Rewrite the system of equations in matrix form:
2 3 −1
5−2 4
3 1 −2
x
y
z
=
4
7
1
Step 2: Use row operations to simplify the augmented matrix:
2 3 −1|4
5−2 4 |7
3 1 −2|1
Step 3: Perform row operations to obtain the following row-echelon form:
100|3
010|2
001| −2
Step 4: Therefore, the solution to the system of equations is x= 3, y= 2,
and z=−2.
34
