MATH 112 - TECHNICAL
MATHEMATICS - Applications to
systems of linear equations
Question Bank - Set 5
Liberty University
Question 1
Question
Solve the following system of linear equations:
2x−y+z= 5
4x+y+ 3z= 6
3x−2y+ 2z=−1
Solution
Step 1: Write the system of equations in matrix form.
2−1 1
413
3−2 2
x
y
z
=
5
6
−1
Step 2: Use Gaussian elimination to simplify the augmented matrix.
2−1 1 |5
413|6
3−2 2 | −1
1−0.5 0.5|2.5
0 3 1 | −2
0−3.5 1.5| −9.5
Step 3: Perform further row operations to simplify the augmented matrix.
1−0.5 0.5|2.5
011/3| −2/3
0 0 2 | −4
Step 4: Use back-substitution to find the values of x,y, and z. From the
last row, we have 2z=−4, so z=−2. Substitute z=−2 back into the second
row:
y+1
3(−2) = −2
3⇒y−2
3=−2
3⇒y= 0
Finally, substitute y= 0 and z=−2 back into the first row:
x−1
2(0) + 1(−2) = 2.5⇒x−2=2.5⇒x= 4.5
Therefore, the solution to the system of linear equations is x= 4.5, y= 0,
and z=−2.
Question 2
Question
Consider the system of linear equations:
x−y+ 2z= 5
2x+ 3y−z= 7
3x+ 2y+z= 4
Determine if the system has a unique solution, infinitely many solutions, or no
solution. If a solution exists, find the solution.
Solution
Step 1: Write the system of equations in augmented matrix form:
1−125
2 3 −1 7
3 2 1 4
Step 2: Perform row operations to simplify the augmented matrix. Subtract
2 times the first row from the second row, and subtract 3 times the first row
from the third row.
1−1 2 5
0 5 −5−3
0 5 −5−11
2
Step 3: Subtract the second row from the third row to eliminate the coeffi-
cients of y. This results in two identical rows which means there are infinitely
many solutions.
1−1 2 5
0 5 −5−3
0 0 0 −8
Step 4: Interpret the augmented matrix back into equations. The last row
corresponds to the equation 0 = −8, which is not true. Hence, the system of
equations has no solution.
Therefore, the given system of equations has no solution.
Question 3
Question
Solve the following system of linear equations:
2x−y+ 3z= 4
3x+ 2y−z= 2
x+ 4y+z= 1
Solution
Step 1: Rewrite the system of equations in matrix form. Let Abe the matrix
of coefficients, Xbe the matrix of variables, and Bbe the matrix of constants.
A=
2−1 3
3 2 −1
1 4 1
, X =
x
y
z
,and B=
4
2
1
Step 2: Calculate the determinant of the matrix A, denoted by |A|.
|A|=
2−1 3
3 2 −1
1 4 1
= 2(2×1−(−1)×4)−(−1)(3×1−1×4)+3(3×4−2×1) = 2(2+4)+1(3−4)+3(12−2) = 12
Step 3: Calculate the determinants of the matrices formed by replacing
columns of Awith Bto find x,y, and z.
|Ax|=
4−1 3
2 2 −1
1 4 1
= 4(2 ×1−(−1) ×4) −(−1)(2 ×1−1×4) + 3(2 ×4−2×1) = 48
|Ay|=
2 4 3
3 2 −1
1 1 1
= 2(2 ×1−4×1) −4(3 ×1−1×1) + 3(3 ×1−2×1) = −13
|Az|=
2−1 4
322
141
= 2(2 ×1−(−1) ×4) −(−1)(3 ×1−1×4) + 4(3 ×2−2×1) = −20
3
Step 4: Calculate x=|Ax|
|A|,y=|Ay|
|A|, and z=|Az|
|A|.
x=48
12 = 4,
y=−13
12 =−13
12,
z=−20
12 =−5
3.
Therefore, the solution to the system of linear equations is x= 4, y=−13
12 ,
and z=−5
3.
Question 4
Question
Solve the system of linear equations:
2x+ 3y−z= 7
3x−2y+ 2z=−5
x+y+z= 1
Solution
Step 1: Write the system of equations in matrix form:
2 3 −1
3−2 2
1 1 1
x
y
z
=
7
−5
1
Step 2: Use row operations to row reduce the augmented matrix:
2 3 −1 7
3−2 2 −5
1 1 1 1
Step 3: Apply row operations to get the matrix in row-echelon form:
1 1 1 1
0−5 5 −8
0 0 −4−4
Step 4: Solve for the variables starting from the last row:
z= 1
Step 5: Substitute z= 1 back into the second row to solve for y:
−5y+ 5 = −8
4
y= 1
Step 6: Substitute y= 1 and z= 1 into the first row to solve for x:
x+ 1 −1=1
x= 1
Step 7: Therefore, the solution to the system of linear equations is x= 1,
y= 1, z= 1.
Question 5
Question
Consider the system of linear equations below:
2x+ 4y+ 6z= 10
3x+ 5y+ 7z= 11
4x+ 6y+ 8z= 12
Find the solution to the system of linear equations, if it exists.
Solution
Step 1: We start by expressing the system of equations in matrix form:
246
357
468
x
y
z
=
10
11
12
Step 2: Next, we form the augmented matrix by combining the coefficient
matrix and the constants:
246|10
357|11
468|12
Step 3: We can perform row operations to row-reduce the augmented matrix:
246|10
357|11
468|12
⇒
1 2 3 |5
0−1−2| −4
0−2−4| −8
⇒
101|1
012|4
000|0
Step 4: The row-reduced form of the augmented matrix reveals that the
system is inconsistent. Thus, the system of linear equations has no solution.
5
Question 6
Question
Solve the following system of linear equations:
2x−3y+z= 5
x+ 2y−z=−4
3x+y+ 4z= 10
Solution
Step 1: Write the system of equations in matrix form:
2−3 1
1 2 −1
3 1 4
x
y
z
=
5
−4
10
Step 2: Use Gaussian Elimination to solve the system. First, let’s perform
row operations to put the augmented matrix in row-echelon form.
Step 3: Multiply the first row by 1
2and subtract it from the second row:
2−3 1
07
2−3
2
3 1 4
Step 4: Multiply the first row by 3
2and subtract it from the third row:
2−3 1
07
2−3
2
011
2
5
2
Step 5: Multiply the second row by 2
7and subtract it from the third row:
2−3 1
07
2−3
2
0 0 4
Step 6: Now, back substitute to find the values of z,y, and x.
From the last row, we have 4z= 4 =⇒z= 1.
Substitute z= 1 back into the second row:
7
2y−3
2=11
2=⇒y= 2.
Substitute y= 2 and z= 1 back into the first row:
2x−3(2) + 1 = 5 =⇒2x= 10 =⇒x= 5.
Therefore, the solution to the system of equations is x= 5, y= 2, and z= 1.
6
Question 7
Question
Consider the following system of linear equations:
3x+ 2y−z= 4
x−y+ 2z=−1
2x+y+ 3z= 10
Determine if the system has a unique solution, infinitely many solutions, or no
solution.
Solution
Step 1: Write the system of linear equations in matrix form:
3 2 −1
1−1 2
2 1 3
x
y
z
=
4
−1
10
Step 2: Find the determinant of the coefficient matrix.
det
3 2 −1
1−1 2
2 1 3
= 3 ·(−1) ·3+2·2·2+(−1) ·1·1 = −3+8−1=4
Since the determinant is not zero, the system has a unique solution.
Therefore, the system of linear equations has a unique solution.
Question 8
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 4y−2z= 3
3x−2y+ 5z= 1
Solution
Step 1: Write the system of linear equations in matrix form.
2−3 1
1 4 −2
3−2 5
x
y
z
=
4
3
1
7
Step 2: Use Gaussian elimination to solve the system of equations.
2−3 1 |4
1 4 −2|3
3−2 5 |1
Step 3: Perform row operations to simplify the matrix.
1 4 −2|3
011
2−3
2|5
2
0−14 11 | −8
Step 4: Continue row operations to simplify the matrix.
1 4 −2|3
0 1 −3
11 |5
11
0−14 11 | −8
Step 5: Perform more row operations to get the identity matrix on the left.
1 0 6
11 |7
11
0 1 −3
11 |5
11
0 0 0 |1
Step 6: Since the last row indicates 0 = 1, there is no solution to this
system of equations. Thus, the system is inconsistent.
Question 9
Question
Consider the following system of linear equations:
2x+ 4y−z= 6
3x+ 2y+z= 5
x+ 2y+ 3z= 7
Determine if this system has a unique solution, infinite solutions, or no solution.
If the system has a unique solution, solve for x,y, and z.
Solution
Step 1: Write the augmented matrix corresponding to the system of equations:
2 4 −1|6
3 2 1 |5
1 2 3 |7
Step 2: Perform row operations to row-reduce the matrix:
8
We will first use the elementary row operation −2R1 + R2→R2.
2 4 −1|6
−1−6 3 | −7
123|7
Next, we will use the elementary row operation −R1 + R3→R3.
2 4 −1|6
−1−6 3 | −7
−1−2 4 |1
Step 3: Continue row reducing the matrix:
Using the elementary row operation R1/2→R1.
1 2 −1/2|3
−1−6 3 | −7
−1−2 4 |1
Next, use the elementary row operations R1 + R2→R2 and R1 + R3→R3.
1 2 −1/2|3
0−4 5/2| −4
0 0 7/2|4
Step 4: Interpret the row-reduced matrix:
The row-reduced matrix has a pivot in each row, which means the system
has a unique solution.
Step 5: Solve the system of equations:
From the row-reduced matrix, we get the equations:
x+ 2y−1
2z= 3
−4y+5
2z=−4
7
2z= 4
Solving the last equation, we find z=8
7.
Substitute z=8
7back into the second equation to solve for y:
−4y+5
2·8
7=−4
Solving, we find y=11
14 .
Finally, substitute z=8
7and y=11
14 into the first equation to solve for x:
x+ 2 ·11
14 −1
2·8
7= 3
Solving, we find x=1
7.
Therefore, the solution to the system of equations is x=1
7,y=11
14 ,z=8
7.
9
Question 10
Question
Consider the system of linear equations:
3x−2y+z= 5
x+y−2z=−3
2x−3y−z= 4
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 system in augmented matrix form:
3−2 1 |5
1 1 −2| −3
2−3−1|4
Step 2: Perform row operations to put the matrix in row-echelon form:
1 1 −2| −3
0−5 7 |14
0 0 1 |1
Step 3: Transform the row-echelon form into reduced row-echelon form:
1 0 −1| −4
0 1 −1.4| −2.8
0 0 1 |1
Step 4: Convert the reduced row-echelon form back into equations:
x−z=−4
y−1.4z=−2.8
z= 1
Step 5: Use back substitution to find the values of x,y, and z: From the
third equation, z= 1. Substitute z= 1 into the second equation:
y−1.4(1) = −2.8
y−1.4 = −2.8
y=−1.4
Substitute y=−1.4 and z= 1 into the first equation:
x−1 = −4
x=−3
Therefore, the system has a unique solution: x=−3, y=−1.4, and z= 1.
10
Question 11
Question
Consider the system of linear equations:
2x+ 3y−z= 4
4x−y+ 2z= 3
3x+ 2y−5z= 5
Determine the solution to this system of linear equations.
Solution
Step 1: Write the system of equations in matrix form. Let Abe the coefficient
matrix, Xbe the matrix of variables, and Bbe the constant matrix. Then the
system of equations can be written as AX =B, where:
A=
2 3 −1
4−1 2
3 2 −5
, X =
x
y
z
,and B=
4
3
5
Step 2: Find the inverse of matrix Aby calculating inv(A).
inv(A) = 1
det(A)·adj(A)
where det(A) is the determinant of matrix Aand adj(A) is the adjugate of
matrix A.
Step 3: Calculate the determinant of matrix A.
det(A) =
2 3 −1
4−1 2
3 2 −5
Step 4: Calculate the adjugate of matrix A.
adj(A) =
−9−13 7
−26 −16 6
−6−8 4
Step 5: Compute the inverse of matrix A.
inv(A) = 1
−49
−9−13 7
−26 −16 6
−6−8 4
Step 6: Find the solution matrix Xby solving the equation X= inv(A)·B.
X=
−9−13 7
−26 −16 6
−6−8 4
·
4
3
5
Step 7: Perform the matrix multiplication to find the solution to the system
of linear equations.
11
Question 12
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 2y−z=−1
4x−y+ 2z= 7
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
1 2 −1
4−1 2
x
y
z
=
4
−1
7
Step 2: Use row operations to get the augmented matrix into row-echelon
form.
2−3 1 4
1 2 −1−1
4−1 2 7
→
1 2 −1−1
0−11
2
7
2
0−5 6 11
Step 3: Continue row operations to get the augmented matrix into reduced
row-echelon form.
1 2 −1−1
0−11
2
7
2
0−5 6 11
→
1 0 3
22
0 1 −1
2−7
2
0 0 11
29
Step 4: Solve for the variables using back substitution. From the last row,
we have 11
2z= 9 ⇒z=18
11
Substitute z=18
11 into the second row to solve for y:y−1
218
11 =−7
2
y−9
11 =−7
2
y=−16
11
Substitute y=−16
11 and z=18
11 into the first row to solve for x:x+ 2 −16
11 +
18
11 =−1
x−32
11 +18
11 =−1
x−14
11 =−1
x=−3
11
Step 5: Thus, the solution to the system of linear equations is:
x=−3
11, y =−16
11, z =18
11
12
Question 13
Question
Solve the following system of linear equations for xand y:
(2x−3y= 7
4x+ 5y= 5
Solution
Step 1: We can solve this system using the method of substitution or elimination.
Let’s use the elimination method for this example.
Step 2: Multiply the first equation by 4 and the second equation by 2 to
make the coefficients of xin both equations equal:
(8x−12y= 28
8x+ 10y= 10
Step 3: Subtract the second new equation from the first new equation:
(8x−12y)−(8x+ 10y) = 28 −10
−22y= 18
y=−9
11
Step 4: Substitute the value of yback into the first original equation to solve
for x:
2x−3(−9
11)=7
Step 5: Simplify and solve for x:
2x+27
11 = 7
2x=77
11 −27
11 =50
11
x=25
11
Step 6: Therefore, the solution to the system of linear equations is x=25
11
and y=−9
11 .
13
Question 14
Question
Solve the following system of linear equations using the method of your choice:
2x−y+z= 3
x+ 3y+ 2z= 9
3x+ 2y+ 4z= 17
Solution
Step 1: We’ll solve this system of equations using the method of substitution.
Step 2: From the first equation, we can express yin terms of xand z:
y= 2x+z−3
Step 3: Substitute this expression for yinto the second equation:
x+ 3(2x+z−3) + 2z= 9
Step 4: Simplify the equation:
x+ 6x+ 3z−9+2z= 9
7x+ 5z= 18
Step 5: Similarly, we substitute the expression for yinto the third equation:
3x+ 2(2x+z−3) + 4z= 17
Step 6: Simplify the equation:
3x+ 4x+ 2z−6+4z= 17
7x+ 6z= 23
Step 7: Now, we have the system of equations:
(7x+ 5z= 18
7x+ 6z= 23
Step 8: Solve this system of equations by subtraction method:
(7x+ 6z)−(7x+ 5z) = 23 −18
z= 5
Step 9: Substitute z= 5 back into 7x+ 5z= 18:
7x+ 5(5) = 18
14
7x+ 25 = 18
7x=−7
x=−1
Step 10: Substitute x=−1 and z= 5 back into y= 2x+z−3:
y= 2(−1) + 5 −3
y=−2
Step 11: Therefore, the solution to the system of equations is:
x=−1, y =−2, z = 5
Question 15
Question
Solve the following system of linear equations using the method of substitution:
3x+ 2y= 11
x−4y=−3
Solution
1. We start by solving the second equation for xin terms of y:
x= 4y−3
2. Next, we substitute xin the first equation with 4y−3:
3(4y−3) + 2y= 11
3. Simplifying the equation:
12y−9+2y= 11 =⇒14y−9 = 11 =⇒14y= 20 =⇒y=20
14 =10
7
4. Now that we have found y, we can substitute it back into the equation
x= 4y−3 to find x:
x= 4 10
7−3 = 40
7−21
7=19
7
5. Therefore, the solution to the system of linear equations is x=19
7and
y=10
7.
15
Question 16
Question
Consider the following system of linear equations:
2x−3y+z= 4
x+ 4y−2z=−3
3x−2y−z= 5
Determine whether the system has a unique solution, infinitely many solu-
tions, or no solution. If there is a unique solution, find the values of x,y, and
z.
Solution
Step 1: Write the augmented matrix for the system of equations:
2−3 1 |4
1 4 −2| −3
3−2−1|5
Step 2: Use row operations to simplify the matrix.
2−3 1 |4
1 4 −2| −3
3−2−1|5
→
1 4 −2| −3
2−3 1 |4
3−2−1|5
Step 3: Use row operations to further simplify the matrix.
1 4 −2| −3
2−3 1 |4
3−2−1|5
→
1 4 −2| −3
0−11 5 |10
0 14 −5|14
Step 4: Continue with row operations to reach row-echelon form.
1 4 −2| −3
0−11 5 |10
0 14 −5|14
→
1 4 −2| −3
0 1 −5
11 | −10
11
0 0 −15
11 |24
11
Step 5: Solve for z,y, and xin terms of the free variable t.
From the last row of the row-echelon form, we have:
−15
11z=24
11 =⇒z=−24
15 =−8
5
Substitute z=−8
5back into the second row:
y−5
11 =−10
11 =⇒y=−10
11 +5
11 =−5
11
16
Substitute z=−8
5and y=−5
11 back into the first row:
x+ 4 −5
11+ 2 −8
5=−3 =⇒x=−3 + 20
11 +16
5=13
55
Therefore, the system has a unique solution: x=13
55 ,y=−5
11 , and z=−8
5.
Question 17
Question
Solve the system of linear equations:
3x−2y+z= 7
2x+y−z=−1
x+ 4y−2z= 7
Solution
Step 1: Write the system of equations in matrix form (AX = B):
3−2 1
2 1 −1
1 4 −2
x
y
z
=
7
−1
7
Step 2: Apply row operations to row reduce the augmented matrix:
3−2 1 |7
2 1 −1| −1
1 4 −2|7
Step 3: Perform row operations to get the matrix in row-echelon form:
1 4 −2|7
0−7 5 | −15
0 0 0 |0
Step 4: Solve the system of equations using back substitution: From the
third row, we get 0z= 0, which tells us that zis a free variable. Let z=t,
where t∈R.
From the second row, we find −7y+ 5t=−15, so y=5t−15
−7.
From the first row, we have x+ 4 5t−15
−7−2t= 7, which simplifies to
x=22−3t
−7.
Therefore, the solution to the system of equations is:
x=22−3t
−7
y=5t−15
−7
z=t
17
Question 18
Question
Solve the following system of linear equations:
2x+ 3y−z= 5
x−2y+ 2z=−1
3x−y+ 4z= 6
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 −1
1−2 2
3−1 4
x
y
z
=
5
−1
6
Step 2: Using row operations, let’s convert the matrix into row-echelon form.
2 3 −1|5
1−2 2 | −1
3−1 4 |6
Step 3: Multiply the first row by 1
2and subtract the result from the second
row:
2 3 −1|5
0−7
2
5
2| −11
2
3−1 4 |6
Step 4: Multiply the first row by 3
2and subtract the result from the third
row:
2 3 −1|5
0−7
2
5
2| −11
2
0−10
2
11
2| −3
2
Step 5: Multiply the second row by −2
7:
2 3 −1|5
0 1 −5
7|11
7
0−10
2
11
2| −3
2
Step 6: Add 10
7times the second row to the third row:
2 3 −1|5
0 1 −5
7|11
7
0 0 3 |2
Step 7: Solve for z: 3z= 2 ⇒z=2
3.
18
Step 8: Substitute z=2
3back into the second row to solve for y:y−5
72
3=
11
7⇒y= 1.
Step 9: Substitute z=2
3and y= 1 back into the first row to solve for x:
2x+ 3(1) −2
3= 5 ⇒x= 2.
Therefore, the solution to the system of linear equations is x= 2, y= 1, and
z=2
3.
Question 19
Question
Solve the system of linear equations using the method of substitution:
(3x+ 2y= 11
2x−5y=−13
Solution
Step 1: Solve the first equation for xin terms of y.
3x+ 2y= 11
3x= 11 −2y
x=11 −2y
3
Step 2: Substitute the expression for xinto the second equation.
211 −2y
3−5y=−13
22 −4y
3−5y=−13
22 −4y−15y=−39
22 −19y=−39
−19y=−61
y=−61
−19
y=61
19
Step 3: Substitute the value of yback into the expression for xto find its
19
value.
x=11 −2(61
19 )
3
=11 −122
19
3
=209 −122
57
=87
57
=29
19
Therefore, the solution to the system of linear equations is x=29
19 and
y=61
19 .
Question 20
Question
Solve the following system of linear equations using the substitution method:
(3x−2y= 4
5x+y=−7
Solution
Step 1: Solve the second equation for yin terms of x.
5x+y=−7
y=−5x−7
Step 2: Substitute y=−5x−7 into the first equation.
3x−2(−5x−7) = 4
3x+ 10x+ 14 = 4
13x+ 14 = 4
13x=−10
x=−10
13
20
Step 3: Substitute x=−10
13 back into y=−5x−7 to find y.
y=−5−10
13−7
y=50
13 −7
y=50 −91
13
y=−41
13
Step 4: The solution to the system of equations is x=−10
13 and y=−41
13 .
Question 21
Question
Solve the following system of linear equations using the method of substitution:
(3x−2y= 7
5x+ 4y= 11
Solution
Step 1: Solve the first equation for xin terms of y:
3x−2y= 7
3x= 2y+ 7
x=2y+ 7
3
Step 2: Substitute x=2y+7
3into the second equation:
52y+ 7
3+ 4y= 11
10y+ 35
3+ 4y= 11
10y+ 35 + 12y= 33
22y=−2
y=−1
11
21
Step 3: Substitute y=−1
11 back into x=2y+7
3:
x=2(−1
11 )+7
3
x=−2 + 77
33
x=75
33
x≈2.27
Therefore, the solution to the system of linear equations is x≈2.27 and
y=−1
11 .
Question 22
Question
Solve the following system of linear equations:
2x−y+z= 3
x+y−z= 0
4x+ 3y+z= 7
Solution
Step 1: Express the system of equations as an augmented matrix:
2−1 1 3
1 1 −1 0
4 3 1 7
Step 2: Perform row operations to simplify the matrix. Multiply the second
row by 2 and subtract the first row from it, and multiply the third row by -2
and add it to the first row.
2−1 1 3
0 3 −3−6
0 1 −1 1
Step 3: Multiply the second row by 1
3to simplify the matrix further.
2−1 1 3
0 1 −1−2
0 1 −1 1
Step 4: Subtract the second row from the third row to get a new third row.
This will lead to a row of zeroes which indicates that the system is dependent.
2−1 1 3
0 1 −1−2
0 0 0 3
22
Step 5: Since the system is dependent and the third row shows a contradic-
tion (0 = 3), there is no solution to the system of equations. Thus, the system
is inconsistent.
Question 23
Question
Solve the following system of linear equations:
2x−3y+ 4z= 7
x+y−z= 2
4x+ 5y−6z= 5
Solution
Step 1: Rewrite the system of equations in matrix form:
2−3 4
1 1 −1
4 5 −6
x
y
z
=
7
2
5
Step 2: Use row operations to reduce the augmented matrix to row-echelon
form:
2−3 4 |7
1 1 −1|2
4 5 −6|5
→
1 1 −1|2
0−5 6 |3
0 1 −2| −3
Step 3: Continue using row operations to further reduce the augmented
matrix:
1 1 −1|2
0−5 6 |3
0 1 −2| −3
→
1 1 −1|2
0 1 −2| −3
0−5 6 |3
→
1 0 1 | −1
0 1 −2| −3
0 0 −4| −12
Step 4: Express the system of equations in parametric form:
x=−1 + z
y=−3+2z
zis free
Step 5: Therefore, the solution to the system of linear equations is:
x=−1 + z
y=−3+2z
z∈R
23
Question 24
Question
Consider the following system of linear equations:
x+ 2y−z= 7
2x−y+z= 4
3x+ 2y+ 2z= 10
Find the solution to this system of equations.
Solution
Step 1: Write the augmented matrix for the system of equations:
1 2 −1|7
2−1 1 |4
3 2 2 |10
Step 2: Perform row operations to put the matrix in row-echelon form:
1 2 −1|7
0−5 3 | −10
0−4 5 | −11
Step 3: Continue row operations to put the matrix in reduced row-echelon
form:
100|1
010| −1
001|2
Step 4: Interpret the matrix to find the solution to the system of equations:
The solution to the system of equations is x= 1, y =−1, z = 2.
Question 25
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
3x−2y+ 2z= 3
x+y−z= 0
24
Solution
Step 1: We will begin by expressing the system of equations in matrix form.
2 3 −1
3−2 2
1 1 −1
x
y
z
=
1
3
0
Step 2: Next, we will row reduce the augmented matrix to row-echelon form
to solve for x,y, and z.
2 3 −1 1
3−2 2 3
1 1 −1 0
→
1 0 −3−7
0 1 1 4
0 0 0 0
Step 3: From the row-echelon form, we obtain the equations:
(x−3z=−7
y+z= 4
Step 4: Solving the above equations, we find x=−7+3zand y= 4 −z.
Step 5: Therefore, the solution to the system of equations is:
x=−7+3z
y= 4 −z
zis free
Question 26
Question
Solve the following system of linear equations by the method of your choice:
2x−3y+z= 1
3x+ 4y−2z= 3
x+y+z= 0
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−3 1 1
3 4 −2 3
1 1 1 0
Step 2: Perform row operations to get zeros below the diagonal in the first
column:
2−3 1 1
3 4 −2 3
1 1 1 0
→
2−3 1 1
017
2−7
2
5
2
1 1 1 0
25
Step 3: Continue with row operations to simplify the matrix further:
2−3 1 1
017
2−7
2
5
2
1 1 1 0
→
2−3 1 1
017
2−7
2
5
2
05
2
1
2−1
2
Step 4: Further simplify the matrix by getting zeros below the diagonal in
the second column:
2−3 1 1
017
2−7
2
5
2
05
2
1
2−1
2
→
2−3 1 1
017
2−7
2
5
2
0 0 4 −4
Step 5: Solve for zfrom the reduced matrix:
z=−1
Step 6: Substitute z=−1 back into the second row to solve for y:
17
2y−7
2(−1) = 5
2
y=3
2
Step 7: Finally, substitute y=3
2and z=−1 back into the first row to solve
for x:
2x−3(3
2)+(−1) = 1
x=7
2
Therefore, the solution to the system of linear equations is x=7
2,y=3
2,
z=−1.
Question 27
Question
Solve the following system of linear equations:
2x+ 3y−z= 7
x−2y+ 3z=−2
3x+ 2y+ 2z= 10
26
Solution
Step 1: Write down the augmented matrix for the system of linear equations.
2 3 −1 7
1−2 3 −2
3 2 2 10
Step 2: Perform row operations to simplify the matrix.
2 3 −1 7
1−2 3 −2
3 2 2 10
→
1−2 3 −2
2 3 −1 7
3 2 2 10
→
1−2 3 −2
0 7 −7 11
0 8 −7 16
Step 3: Continue to simplify the matrix.
1−2 3 −2
0 7 −7 11
0 8 −7 16
→
1−2 3 −2
0 1 −111
7
0 8 −7 16
→
1−2 3 −2
0 1 −111
7
0 0 1 −5
7
Step 4: Convert the matrix back to equations and solve for x,y, and z.
From the last row of the matrix, we get:
z=−5
7
From the second row of the matrix, we get:
y−z=11
7
y−−5
7=11
7
y+5
7=11
7
y=6
7
From the first row of the matrix, we get:
x−2y+ 3z=−2
x−26
7+ 3 −5
7=−2
x−12
7−15
7=−2
x=−2 + 27
7
x=8
7
Therefore, the solution to the system of linear equations is:
x=8
7, y =6
7, z =−5
7
27
Question 28
Question
Consider the system of linear equations:
4x−y+ 2z= 3
3x+ 2y−z= 7
x+ 3y+z=−2
Determine if the system has a unique solution, no solution, or infinitely many
solutions. If a unique solution exists, find the values of x,y, and z.
Solution
Step 1: Write the system of equations in augmented matrix form:
4−1 2 |3
3 2 −1|7
1 3 1 | −2
Step 2: Perform row operations to simplify the augmented matrix:
4−1 2 |3
3 2 −1|7
1 3 1 | −2
R2−3
4R1
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
1 3 1 | −2
R3−1
4R1
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
013
4
1
2| −11
4
R3−13
11 R2
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
0 0 3 | − 20
3
Step 3: From the row-echelon form, we can see that the last row corresponds
to the equation 0x+ 0y+ 3z=−20
3, which simplifies to z=−20
9.
Step 4: Substitute z=−20
3back into the second-to-last row in the aug-
mented matrix: 11
4y−11
4(−20
3) = 17
4
Solving this equation gives y= 1.
Step 5: Substitute y= 1 and z=−20
3back into the first row in the aug-
mented matrix:
4x−1 + 2(−20
3)=3
Solving this equation gives x= 5.
Hence, the system of equations has a unique solution: x= 5, y= 1, and
z=−20
3.
28
Question 29
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−y+ 2z=−3
2x+y−3z=−1
Solution
Step 1: Write the system of equations in matrix form.
3 2 −1
1−1 2
2 1 −3
x
y
z
=
4
−3
−1
Step 2: Use an augmented matrix to apply row operations and solve for x,
y, and z.
3 2 −1|4
1−1 2 | −3
2 1 −3| −1
Step 3: Perform row operations to get a row-echelon form.
3 2 −1|4
0−1.67 2.33 | −4.33
0 0 −4| −9
Step 4: Solve for zusing the third equation (−4z=−9), giving z=9
4=
−2.25.
Step 5: Substitute zinto the second equation to solve for y.
−y+ 2(−2.25) = −4.33
−y−4.5 = −4.33
−y= 0.17
y=−0.17
Step 6: Substitute zand yback into the first equation to solve for x.
3x+ 2(−0.17) −(−2.25) = 4
3x−0.34 + 2.25 = 4
3x+ 1.91 = 4
3x= 2.09
x≈0.70
Therefore, the solution to the system of linear equations is x≈0.70, y=
−0.17, and z=−2.25.
29
Question 30
Question
Solve the system of linear equations:
2x−3y+z= 6
3x+ 2y−2z= 2
x+y+ 3z= 4
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 6
3 2 −2 2
1 1 3 4
Step 2: Perform row operations to row-reduce the matrix.
1 1 3 4
05
2−13
2−10
0−2−5 2
Step 3: Further row operations yield:
1 1 3 4
0 1 −13
5−4
0 0 −27
5−18
Step 4: Solve for zfrom the third row:
−27
5z=−18 =⇒z=90
27 = 3
Step 5: Substitute z= 3 into the second row:
y−13
5(3) = −4 =⇒y=−4 + 39
5=−1
5
Step 6: Substitute y=−1
5and z= 3 into the first row:
x−1
5+ 9 = 4 =⇒x= 4 + 1
5−9 = −31
5
Step 7: The solution to the system of equations is x=−31
5,y=−1
5,z= 3.
30
Question 31
Question
Solve the following system of linear equations using matrix methods:
2x−3y+z= 4
x+ 4y−2z=−3
3x−2y+ 3z= 10
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 |4
1 4 −2| −3
3−2 3 |10
Step 2: Perform row operations to get the matrix into reduced row-echelon
form.
100|2
010| −1
001|1
Step 3: Write the solution as a set of equations:
x= 2
y=−1
z= 1
Therefore, the solution to the system of linear equations is:
x= 2, y =−1, z = 1
Question 32
Question
Let’s consider a system of linear equations:
2x−y+ 3z= 4
x+ 2y−z= 1
3x−2y+ 4z= 7
Determine if the system has a unique solution, infinitely many solutions, or no
solution.
31
Solution
Step 1: We can represent the system of linear equations in matrix form as
Ax=bwhere
A=
2−1 3
1 2 −1
3−2 4
,x=
x
y
z
,and b=
4
1
7
.
Step 2: To determine the solution of the system, we find the determinant
of matrix A. If |A| = 0, then the system has a unique solution. If |A|= 0 and
rank(A|b) = rank(A), then the system has infinitely many solutions. If |A|= 0
and rank(A|b)= rank(A), then the system has no solution.
Step 3: Calculate the determinant of matrix A:
2−1 3
1 2 −1
3−2 4
= 2(2·4−(−1)(−2))−(−1)(1·4−(−1)(3))+3(1·−2−2·3) = 28−1−15 = 12.
Since |A|= 12 = 0, the system has a unique solution.
Question 33
Question
Solve the following system of linear equations:
3x−2y+z= 7
x+ 4y−2z=−6
2x−3y+ 2z= 8
Solution
Step 1: Write the system of equations in matrix form. Let
A=
3−2 1
1 4 −2
2−3 2
, X =
x
y
z
, B =
7
−6
8
.
The system of equations can then be written as AX =B.
Step 2: Use Gaussian elimination to row-reduce the augmented matrix [A|B].
[ccc|c]3 −2 1 7
1 4 −2−6
2−3 2 8
→
[ccc|c]1 4 −2−6
0−5 3 15
0−11 5 20
→
[ccc|c]1 4 −2−6
0−5 3 15
0 0 7 85
32
Step 3: Perform back substitution to find the values of x,y, and z. From
the row-reduced form, we have
x+ 4y−2z=−6
−5y+ 3z= 15
7z= 85
Solving the equations, we find z=85
7. Substitute zback to solve for yand x.
Therefore, y= 4 and x=−5.
Therefore, the solution to the system of equations is x=−5, y= 4, and
z=85
7.
Question 34
Question
Solve the system of linear equations using the Gauss-Jordan elimination method:
2x+ 3y−z= 1
4x+ 2y+ 3z= 4
3x−y+ 2z= 5
Solution
Step 1: Write the augmented matrix for the system of linear equations:
2 3 −1|1
4 2 3 |4
3−1 2 |5
Step 2: Perform row operations to get zeros below the diagonal:
2 3 −1|1
0−4 5 |2
0−10 5 |4
Step 3: Perform row operations to get zeros above the diagonal:
2 3 −1|1
0−4 5 |2
0 0 10 |6
Step 4: Perform row operations to obtain the reduced row-echelon form:
1 0 7
10 |3
5
0 1 −5
8| −1
4
0 0 1 |3
5
33
Step 5: Write the solution back as a system of equations:
x=3
5−7
10z
y=−1
4+5
8z
z=3
5
So, the solution to the system of linear equations is:
x=1
10
y=−1
4
z=3
5
Question 35
Question
Consider the following system of linear equations:
2x+ 3y−z= 7
x−y+ 2z=−1
3x+ 2y+ 4z= 15
Find the solution to the system of equations, or determine if no solution exists.
Solution
Step 1: Write the system of equations in matrix form, Ax=b:
2 3 −1
1−1 2
3 2 4
x
y
z
=
7
−1
15
Step 2: Use Gaussian elimination to row reduce the augmented matrix [A|b].
2 3 −1 7
1−1 2 −1
3 2 4 15
Step 3: Perform row operations to obtain row-echelon form.
1−1 2 −1
0 2.5−2.5 6.5
0 5.5−7 16
34
Question 7
Question
Consider the following system of linear equations:
3x+ 2y−z= 4
x−y+ 2z=−1
2x+y+ 3z= 10
Determine if the system has a unique solution, infinitely many solutions, or no
solution.
Solution
Step 1: Write the system of linear equations in matrix form:
3 2 −1
1−1 2
2 1 3
x
y
z
=
4
−1
10
Step 2: Find the determinant of the coefficient matrix.
det
3 2 −1
1−1 2
2 1 3
= 3 ·(−1) ·3+2·2·2+(−1) ·1·1 = −3+8−1=4
Since the determinant is not zero, the system has a unique solution.
Therefore, the system of linear equations has a unique solution.
Question 8
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 4y−2z= 3
3x−2y+ 5z= 1
Solution
Step 1: Write the system of linear equations in matrix form.
2−3 1
1 4 −2
3−2 5
x
y
z
=
4
3
1
7
Step 2: Use Gaussian elimination to solve the system of equations.
2−3 1 |4
1 4 −2|3
3−2 5 |1
Step 3: Perform row operations to simplify the matrix.
1 4 −2|3
011
2−3
2|5
2
0−14 11 | −8
Step 4: Continue row operations to simplify the matrix.
1 4 −2|3
0 1 −3
11 |5
11
0−14 11 | −8
Step 5: Perform more row operations to get the identity matrix on the left.
1 0 6
11 |7
11
0 1 −3
11 |5
11
0 0 0 |1
Step 6: Since the last row indicates 0 = 1, there is no solution to this
system of equations. Thus, the system is inconsistent.
Question 9
Question
Consider the following system of linear equations:
2x+ 4y−z= 6
3x+ 2y+z= 5
x+ 2y+ 3z= 7
Determine if this system has a unique solution, infinite solutions, or no solution.
If the system has a unique solution, solve for x,y, and z.
Solution
Step 1: Write the augmented matrix corresponding to the system of equations:
2 4 −1|6
3 2 1 |5
1 2 3 |7
Step 2: Perform row operations to row-reduce the matrix:
8
We will first use the elementary row operation −2R1 + R2→R2.
2 4 −1|6
−1−6 3 | −7
123|7
Next, we will use the elementary row operation −R1 + R3→R3.
2 4 −1|6
−1−6 3 | −7
−1−2 4 |1
Step 3: Continue row reducing the matrix:
Using the elementary row operation R1/2→R1.
1 2 −1/2|3
−1−6 3 | −7
−1−2 4 |1
Next, use the elementary row operations R1 + R2→R2 and R1 + R3→R3.
1 2 −1/2|3
0−4 5/2| −4
0 0 7/2|4
Step 4: Interpret the row-reduced matrix:
The row-reduced matrix has a pivot in each row, which means the system
has a unique solution.
Step 5: Solve the system of equations:
From the row-reduced matrix, we get the equations:
x+ 2y−1
2z= 3
−4y+5
2z=−4
7
2z= 4
Solving the last equation, we find z=8
7.
Substitute z=8
7back into the second equation to solve for y:
−4y+5
2·8
7=−4
Solving, we find y=11
14 .
Finally, substitute z=8
7and y=11
14 into the first equation to solve for x:
x+ 2 ·11
14 −1
2·8
7= 3
Solving, we find x=1
7.
Therefore, the solution to the system of equations is x=1
7,y=11
14 ,z=8
7.
9
Question 10
Question
Consider the system of linear equations:
3x−2y+z= 5
x+y−2z=−3
2x−3y−z= 4
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 system in augmented matrix form:
3−2 1 |5
1 1 −2| −3
2−3−1|4
Step 2: Perform row operations to put the matrix in row-echelon form:
1 1 −2| −3
0−5 7 |14
0 0 1 |1
Step 3: Transform the row-echelon form into reduced row-echelon form:
1 0 −1| −4
0 1 −1.4| −2.8
0 0 1 |1
Step 4: Convert the reduced row-echelon form back into equations:
x−z=−4
y−1.4z=−2.8
z= 1
Step 5: Use back substitution to find the values of x,y, and z: From the
third equation, z= 1. Substitute z= 1 into the second equation:
y−1.4(1) = −2.8
y−1.4 = −2.8
y=−1.4
Substitute y=−1.4 and z= 1 into the first equation:
x−1 = −4
x=−3
Therefore, the system has a unique solution: x=−3, y=−1.4, and z= 1.
10
Question 11
Question
Consider the system of linear equations:
2x+ 3y−z= 4
4x−y+ 2z= 3
3x+ 2y−5z= 5
Determine the solution to this system of linear equations.
Solution
Step 1: Write the system of equations in matrix form. Let Abe the coefficient
matrix, Xbe the matrix of variables, and Bbe the constant matrix. Then the
system of equations can be written as AX =B, where:
A=
2 3 −1
4−1 2
3 2 −5
, X =
x
y
z
,and B=
4
3
5
Step 2: Find the inverse of matrix Aby calculating inv(A).
inv(A) = 1
det(A)·adj(A)
where det(A) is the determinant of matrix Aand adj(A) is the adjugate of
matrix A.
Step 3: Calculate the determinant of matrix A.
det(A) =
2 3 −1
4−1 2
3 2 −5
Step 4: Calculate the adjugate of matrix A.
adj(A) =
−9−13 7
−26 −16 6
−6−8 4
Step 5: Compute the inverse of matrix A.
inv(A) = 1
−49
−9−13 7
−26 −16 6
−6−8 4
Step 6: Find the solution matrix Xby solving the equation X= inv(A)·B.
X=
−9−13 7
−26 −16 6
−6−8 4
·
4
3
5
Step 7: Perform the matrix multiplication to find the solution to the system
of linear equations.
11
Question 12
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 2y−z=−1
4x−y+ 2z= 7
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
1 2 −1
4−1 2
x
y
z
=
4
−1
7
Step 2: Use row operations to get the augmented matrix into row-echelon
form.
2−3 1 4
1 2 −1−1
4−1 2 7
→
1 2 −1−1
0−11
2
7
2
0−5 6 11
Step 3: Continue row operations to get the augmented matrix into reduced
row-echelon form.
1 2 −1−1
0−11
2
7
2
0−5 6 11
→
1 0 3
22
0 1 −1
2−7
2
0 0 11
29
Step 4: Solve for the variables using back substitution. From the last row,
we have 11
2z= 9 ⇒z=18
11
Substitute z=18
11 into the second row to solve for y:y−1
218
11 =−7
2
y−9
11 =−7
2
y=−16
11
Substitute y=−16
11 and z=18
11 into the first row to solve for x:x+ 2 −16
11 +
18
11 =−1
x−32
11 +18
11 =−1
x−14
11 =−1
x=−3
11
Step 5: Thus, the solution to the system of linear equations is:
x=−3
11, y =−16
11, z =18
11
12
Question 13
Question
Solve the following system of linear equations for xand y:
(2x−3y= 7
4x+ 5y= 5
Solution
Step 1: We can solve this system using the method of substitution or elimination.
Let’s use the elimination method for this example.
Step 2: Multiply the first equation by 4 and the second equation by 2 to
make the coefficients of xin both equations equal:
(8x−12y= 28
8x+ 10y= 10
Step 3: Subtract the second new equation from the first new equation:
(8x−12y)−(8x+ 10y) = 28 −10
−22y= 18
y=−9
11
Step 4: Substitute the value of yback into the first original equation to solve
for x:
2x−3(−9
11)=7
Step 5: Simplify and solve for x:
2x+27
11 = 7
2x=77
11 −27
11 =50
11
x=25
11
Step 6: Therefore, the solution to the system of linear equations is x=25
11
and y=−9
11 .
13
Question 14
Question
Solve the following system of linear equations using the method of your choice:
2x−y+z= 3
x+ 3y+ 2z= 9
3x+ 2y+ 4z= 17
Solution
Step 1: We’ll solve this system of equations using the method of substitution.
Step 2: From the first equation, we can express yin terms of xand z:
y= 2x+z−3
Step 3: Substitute this expression for yinto the second equation:
x+ 3(2x+z−3) + 2z= 9
Step 4: Simplify the equation:
x+ 6x+ 3z−9+2z= 9
7x+ 5z= 18
Step 5: Similarly, we substitute the expression for yinto the third equation:
3x+ 2(2x+z−3) + 4z= 17
Step 6: Simplify the equation:
3x+ 4x+ 2z−6+4z= 17
7x+ 6z= 23
Step 7: Now, we have the system of equations:
(7x+ 5z= 18
7x+ 6z= 23
Step 8: Solve this system of equations by subtraction method:
(7x+ 6z)−(7x+ 5z) = 23 −18
z= 5
Step 9: Substitute z= 5 back into 7x+ 5z= 18:
7x+ 5(5) = 18
14
7x+ 25 = 18
7x=−7
x=−1
Step 10: Substitute x=−1 and z= 5 back into y= 2x+z−3:
y= 2(−1) + 5 −3
y=−2
Step 11: Therefore, the solution to the system of equations is:
x=−1, y =−2, z = 5
Question 15
Question
Solve the following system of linear equations using the method of substitution:
3x+ 2y= 11
x−4y=−3
Solution
1. We start by solving the second equation for xin terms of y:
x= 4y−3
2. Next, we substitute xin the first equation with 4y−3:
3(4y−3) + 2y= 11
3. Simplifying the equation:
12y−9+2y= 11 =⇒14y−9 = 11 =⇒14y= 20 =⇒y=20
14 =10
7
4. Now that we have found y, we can substitute it back into the equation
x= 4y−3 to find x:
x= 4 10
7−3 = 40
7−21
7=19
7
5. Therefore, the solution to the system of linear equations is x=19
7and
y=10
7.
15
Question 16
Question
Consider the following system of linear equations:
2x−3y+z= 4
x+ 4y−2z=−3
3x−2y−z= 5
Determine whether the system has a unique solution, infinitely many solu-
tions, or no solution. If there is a unique solution, find the values of x,y, and
z.
Solution
Step 1: Write the augmented matrix for the system of equations:
2−3 1 |4
1 4 −2| −3
3−2−1|5
Step 2: Use row operations to simplify the matrix.
2−3 1 |4
1 4 −2| −3
3−2−1|5
→
1 4 −2| −3
2−3 1 |4
3−2−1|5
Step 3: Use row operations to further simplify the matrix.
1 4 −2| −3
2−3 1 |4
3−2−1|5
→
1 4 −2| −3
0−11 5 |10
0 14 −5|14
Step 4: Continue with row operations to reach row-echelon form.
1 4 −2| −3
0−11 5 |10
0 14 −5|14
→
1 4 −2| −3
0 1 −5
11 | −10
11
0 0 −15
11 |24
11
Step 5: Solve for z,y, and xin terms of the free variable t.
From the last row of the row-echelon form, we have:
−15
11z=24
11 =⇒z=−24
15 =−8
5
Substitute z=−8
5back into the second row:
y−5
11 =−10
11 =⇒y=−10
11 +5
11 =−5
11
16
Substitute z=−8
5and y=−5
11 back into the first row:
x+ 4 −5
11+ 2 −8
5=−3 =⇒x=−3 + 20
11 +16
5=13
55
Therefore, the system has a unique solution: x=13
55 ,y=−5
11 , and z=−8
5.
Question 17
Question
Solve the system of linear equations:
3x−2y+z= 7
2x+y−z=−1
x+ 4y−2z= 7
Solution
Step 1: Write the system of equations in matrix form (AX = B):
3−2 1
2 1 −1
1 4 −2
x
y
z
=
7
−1
7
Step 2: Apply row operations to row reduce the augmented matrix:
3−2 1 |7
2 1 −1| −1
1 4 −2|7
Step 3: Perform row operations to get the matrix in row-echelon form:
1 4 −2|7
0−7 5 | −15
0 0 0 |0
Step 4: Solve the system of equations using back substitution: From the
third row, we get 0z= 0, which tells us that zis a free variable. Let z=t,
where t∈R.
From the second row, we find −7y+ 5t=−15, so y=5t−15
−7.
From the first row, we have x+ 4 5t−15
−7−2t= 7, which simplifies to
x=22−3t
−7.
Therefore, the solution to the system of equations is:
x=22−3t
−7
y=5t−15
−7
z=t
17
Question 18
Question
Solve the following system of linear equations:
2x+ 3y−z= 5
x−2y+ 2z=−1
3x−y+ 4z= 6
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 −1
1−2 2
3−1 4
x
y
z
=
5
−1
6
Step 2: Using row operations, let’s convert the matrix into row-echelon form.
2 3 −1|5
1−2 2 | −1
3−1 4 |6
Step 3: Multiply the first row by 1
2and subtract the result from the second
row:
2 3 −1|5
0−7
2
5
2| −11
2
3−1 4 |6
Step 4: Multiply the first row by 3
2and subtract the result from the third
row:
2 3 −1|5
0−7
2
5
2| −11
2
0−10
2
11
2| −3
2
Step 5: Multiply the second row by −2
7:
2 3 −1|5
0 1 −5
7|11
7
0−10
2
11
2| −3
2
Step 6: Add 10
7times the second row to the third row:
2 3 −1|5
0 1 −5
7|11
7
0 0 3 |2
Step 7: Solve for z: 3z= 2 ⇒z=2
3.
18
Step 8: Substitute z=2
3back into the second row to solve for y:y−5
72
3=
11
7⇒y= 1.
Step 9: Substitute z=2
3and y= 1 back into the first row to solve for x:
2x+ 3(1) −2
3= 5 ⇒x= 2.
Therefore, the solution to the system of linear equations is x= 2, y= 1, and
z=2
3.
Question 19
Question
Solve the system of linear equations using the method of substitution:
(3x+ 2y= 11
2x−5y=−13
Solution
Step 1: Solve the first equation for xin terms of y.
3x+ 2y= 11
3x= 11 −2y
x=11 −2y
3
Step 2: Substitute the expression for xinto the second equation.
211 −2y
3−5y=−13
22 −4y
3−5y=−13
22 −4y−15y=−39
22 −19y=−39
−19y=−61
y=−61
−19
y=61
19
Step 3: Substitute the value of yback into the expression for xto find its
19
value.
x=11 −2(61
19 )
3
=11 −122
19
3
=209 −122
57
=87
57
=29
19
Therefore, the solution to the system of linear equations is x=29
19 and
y=61
19 .
Question 20
Question
Solve the following system of linear equations using the substitution method:
(3x−2y= 4
5x+y=−7
Solution
Step 1: Solve the second equation for yin terms of x.
5x+y=−7
y=−5x−7
Step 2: Substitute y=−5x−7 into the first equation.
3x−2(−5x−7) = 4
3x+ 10x+ 14 = 4
13x+ 14 = 4
13x=−10
x=−10
13
20
Step 3: Substitute x=−10
13 back into y=−5x−7 to find y.
y=−5−10
13−7
y=50
13 −7
y=50 −91
13
y=−41
13
Step 4: The solution to the system of equations is x=−10
13 and y=−41
13 .
Question 21
Question
Solve the following system of linear equations using the method of substitution:
(3x−2y= 7
5x+ 4y= 11
Solution
Step 1: Solve the first equation for xin terms of y:
3x−2y= 7
3x= 2y+ 7
x=2y+ 7
3
Step 2: Substitute x=2y+7
3into the second equation:
52y+ 7
3+ 4y= 11
10y+ 35
3+ 4y= 11
10y+ 35 + 12y= 33
22y=−2
y=−1
11
21
Step 3: Substitute y=−1
11 back into x=2y+7
3:
x=2(−1
11 )+7
3
x=−2 + 77
33
x=75
33
x≈2.27
Therefore, the solution to the system of linear equations is x≈2.27 and
y=−1
11 .
Question 22
Question
Solve the following system of linear equations:
2x−y+z= 3
x+y−z= 0
4x+ 3y+z= 7
Solution
Step 1: Express the system of equations as an augmented matrix:
2−1 1 3
1 1 −1 0
4 3 1 7
Step 2: Perform row operations to simplify the matrix. Multiply the second
row by 2 and subtract the first row from it, and multiply the third row by -2
and add it to the first row.
2−1 1 3
0 3 −3−6
0 1 −1 1
Step 3: Multiply the second row by 1
3to simplify the matrix further.
2−1 1 3
0 1 −1−2
0 1 −1 1
Step 4: Subtract the second row from the third row to get a new third row.
This will lead to a row of zeroes which indicates that the system is dependent.
2−1 1 3
0 1 −1−2
0 0 0 3
22
Step 5: Since the system is dependent and the third row shows a contradic-
tion (0 = 3), there is no solution to the system of equations. Thus, the system
is inconsistent.
Question 23
Question
Solve the following system of linear equations:
2x−3y+ 4z= 7
x+y−z= 2
4x+ 5y−6z= 5
Solution
Step 1: Rewrite the system of equations in matrix form:
2−3 4
1 1 −1
4 5 −6
x
y
z
=
7
2
5
Step 2: Use row operations to reduce the augmented matrix to row-echelon
form:
2−3 4 |7
1 1 −1|2
4 5 −6|5
→
1 1 −1|2
0−5 6 |3
0 1 −2| −3
Step 3: Continue using row operations to further reduce the augmented
matrix:
1 1 −1|2
0−5 6 |3
0 1 −2| −3
→
1 1 −1|2
0 1 −2| −3
0−5 6 |3
→
1 0 1 | −1
0 1 −2| −3
0 0 −4| −12
Step 4: Express the system of equations in parametric form:
x=−1 + z
y=−3+2z
zis free
Step 5: Therefore, the solution to the system of linear equations is:
x=−1 + z
y=−3+2z
z∈R
23
Question 24
Question
Consider the following system of linear equations:
x+ 2y−z= 7
2x−y+z= 4
3x+ 2y+ 2z= 10
Find the solution to this system of equations.
Solution
Step 1: Write the augmented matrix for the system of equations:
1 2 −1|7
2−1 1 |4
3 2 2 |10
Step 2: Perform row operations to put the matrix in row-echelon form:
1 2 −1|7
0−5 3 | −10
0−4 5 | −11
Step 3: Continue row operations to put the matrix in reduced row-echelon
form:
100|1
010| −1
001|2
Step 4: Interpret the matrix to find the solution to the system of equations:
The solution to the system of equations is x= 1, y =−1, z = 2.
Question 25
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
3x−2y+ 2z= 3
x+y−z= 0
24
Solution
Step 1: We will begin by expressing the system of equations in matrix form.
2 3 −1
3−2 2
1 1 −1
x
y
z
=
1
3
0
Step 2: Next, we will row reduce the augmented matrix to row-echelon form
to solve for x,y, and z.
2 3 −1 1
3−2 2 3
1 1 −1 0
→
1 0 −3−7
0 1 1 4
0 0 0 0
Step 3: From the row-echelon form, we obtain the equations:
(x−3z=−7
y+z= 4
Step 4: Solving the above equations, we find x=−7+3zand y= 4 −z.
Step 5: Therefore, the solution to the system of equations is:
x=−7+3z
y= 4 −z
zis free
Question 26
Question
Solve the following system of linear equations by the method of your choice:
2x−3y+z= 1
3x+ 4y−2z= 3
x+y+z= 0
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−3 1 1
3 4 −2 3
1 1 1 0
Step 2: Perform row operations to get zeros below the diagonal in the first
column:
2−3 1 1
3 4 −2 3
1 1 1 0
→
2−3 1 1
017
2−7
2
5
2
1 1 1 0
25
Step 3: Continue with row operations to simplify the matrix further:
2−3 1 1
017
2−7
2
5
2
1 1 1 0
→
2−3 1 1
017
2−7
2
5
2
05
2
1
2−1
2
Step 4: Further simplify the matrix by getting zeros below the diagonal in
the second column:
2−3 1 1
017
2−7
2
5
2
05
2
1
2−1
2
→
2−3 1 1
017
2−7
2
5
2
0 0 4 −4
Step 5: Solve for zfrom the reduced matrix:
z=−1
Step 6: Substitute z=−1 back into the second row to solve for y:
17
2y−7
2(−1) = 5
2
y=3
2
Step 7: Finally, substitute y=3
2and z=−1 back into the first row to solve
for x:
2x−3(3
2)+(−1) = 1
x=7
2
Therefore, the solution to the system of linear equations is x=7
2,y=3
2,
z=−1.
Question 27
Question
Solve the following system of linear equations:
2x+ 3y−z= 7
x−2y+ 3z=−2
3x+ 2y+ 2z= 10
26
Solution
Step 1: Write down the augmented matrix for the system of linear equations.
2 3 −1 7
1−2 3 −2
3 2 2 10
Step 2: Perform row operations to simplify the matrix.
2 3 −1 7
1−2 3 −2
3 2 2 10
→
1−2 3 −2
2 3 −1 7
3 2 2 10
→
1−2 3 −2
0 7 −7 11
0 8 −7 16
Step 3: Continue to simplify the matrix.
1−2 3 −2
0 7 −7 11
0 8 −7 16
→
1−2 3 −2
0 1 −111
7
0 8 −7 16
→
1−2 3 −2
0 1 −111
7
0 0 1 −5
7
Step 4: Convert the matrix back to equations and solve for x,y, and z.
From the last row of the matrix, we get:
z=−5
7
From the second row of the matrix, we get:
y−z=11
7
y−−5
7=11
7
y+5
7=11
7
y=6
7
From the first row of the matrix, we get:
x−2y+ 3z=−2
x−26
7+ 3 −5
7=−2
x−12
7−15
7=−2
x=−2 + 27
7
x=8
7
Therefore, the solution to the system of linear equations is:
x=8
7, y =6
7, z =−5
7
27
Question 28
Question
Consider the system of linear equations:
4x−y+ 2z= 3
3x+ 2y−z= 7
x+ 3y+z=−2
Determine if the system has a unique solution, no solution, or infinitely many
solutions. If a unique solution exists, find the values of x,y, and z.
Solution
Step 1: Write the system of equations in augmented matrix form:
4−1 2 |3
3 2 −1|7
1 3 1 | −2
Step 2: Perform row operations to simplify the augmented matrix:
4−1 2 |3
3 2 −1|7
1 3 1 | −2
R2−3
4R1
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
1 3 1 | −2
R3−1
4R1
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
013
4
1
2| −11
4
R3−13
11 R2
−−−−−−→
4−1 2 |3
011
4−11
4|17
4
0 0 3 | − 20
3
Step 3: From the row-echelon form, we can see that the last row corresponds
to the equation 0x+ 0y+ 3z=−20
3, which simplifies to z=−20
9.
Step 4: Substitute z=−20
3back into the second-to-last row in the aug-
mented matrix: 11
4y−11
4(−20
3) = 17
4
Solving this equation gives y= 1.
Step 5: Substitute y= 1 and z=−20
3back into the first row in the aug-
mented matrix:
4x−1 + 2(−20
3)=3
Solving this equation gives x= 5.
Hence, the system of equations has a unique solution: x= 5, y= 1, and
z=−20
3.
28
Question 29
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−y+ 2z=−3
2x+y−3z=−1
Solution
Step 1: Write the system of equations in matrix form.
3 2 −1
1−1 2
2 1 −3
x
y
z
=
4
−3
−1
Step 2: Use an augmented matrix to apply row operations and solve for x,
y, and z.
3 2 −1|4
1−1 2 | −3
2 1 −3| −1
Step 3: Perform row operations to get a row-echelon form.
3 2 −1|4
0−1.67 2.33 | −4.33
0 0 −4| −9
Step 4: Solve for zusing the third equation (−4z=−9), giving z=9
4=
−2.25.
Step 5: Substitute zinto the second equation to solve for y.
−y+ 2(−2.25) = −4.33
−y−4.5 = −4.33
−y= 0.17
y=−0.17
Step 6: Substitute zand yback into the first equation to solve for x.
3x+ 2(−0.17) −(−2.25) = 4
3x−0.34 + 2.25 = 4
3x+ 1.91 = 4
3x= 2.09
x≈0.70
Therefore, the solution to the system of linear equations is x≈0.70, y=
−0.17, and z=−2.25.
29
Question 30
Question
Solve the system of linear equations:
2x−3y+z= 6
3x+ 2y−2z= 2
x+y+ 3z= 4
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 6
3 2 −2 2
1 1 3 4
Step 2: Perform row operations to row-reduce the matrix.
1 1 3 4
05
2−13
2−10
0−2−5 2
Step 3: Further row operations yield:
1 1 3 4
0 1 −13
5−4
0 0 −27
5−18
Step 4: Solve for zfrom the third row:
−27
5z=−18 =⇒z=90
27 = 3
Step 5: Substitute z= 3 into the second row:
y−13
5(3) = −4 =⇒y=−4 + 39
5=−1
5
Step 6: Substitute y=−1
5and z= 3 into the first row:
x−1
5+ 9 = 4 =⇒x= 4 + 1
5−9 = −31
5
Step 7: The solution to the system of equations is x=−31
5,y=−1
5,z= 3.
30
Question 31
Question
Solve the following system of linear equations using matrix methods:
2x−3y+z= 4
x+ 4y−2z=−3
3x−2y+ 3z= 10
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 |4
1 4 −2| −3
3−2 3 |10
Step 2: Perform row operations to get the matrix into reduced row-echelon
form.
100|2
010| −1
001|1
Step 3: Write the solution as a set of equations:
x= 2
y=−1
z= 1
Therefore, the solution to the system of linear equations is:
x= 2, y =−1, z = 1
Question 32
Question
Let’s consider a system of linear equations:
2x−y+ 3z= 4
x+ 2y−z= 1
3x−2y+ 4z= 7
Determine if the system has a unique solution, infinitely many solutions, or no
solution.
31
Solution
Step 1: We can represent the system of linear equations in matrix form as
Ax=bwhere
A=
2−1 3
1 2 −1
3−2 4
,x=
x
y
z
,and b=
4
1
7
.
Step 2: To determine the solution of the system, we find the determinant
of matrix A. If |A| = 0, then the system has a unique solution. If |A|= 0 and
rank(A|b) = rank(A), then the system has infinitely many solutions. If |A|= 0
and rank(A|b)= rank(A), then the system has no solution.
Step 3: Calculate the determinant of matrix A:
2−1 3
1 2 −1
3−2 4
= 2(2·4−(−1)(−2))−(−1)(1·4−(−1)(3))+3(1·−2−2·3) = 28−1−15 = 12.
Since |A|= 12 = 0, the system has a unique solution.
Question 33
Question
Solve the following system of linear equations:
3x−2y+z= 7
x+ 4y−2z=−6
2x−3y+ 2z= 8
Solution
Step 1: Write the system of equations in matrix form. Let
A=
3−2 1
1 4 −2
2−3 2
, X =
x
y
z
, B =
7
−6
8
.
The system of equations can then be written as AX =B.
Step 2: Use Gaussian elimination to row-reduce the augmented matrix [A|B].
[ccc|c]3 −2 1 7
1 4 −2−6
2−3 2 8
→
[ccc|c]1 4 −2−6
0−5 3 15
0−11 5 20
→
[ccc|c]1 4 −2−6
0−5 3 15
0 0 7 85
32
Step 3: Perform back substitution to find the values of x,y, and z. From
the row-reduced form, we have
x+ 4y−2z=−6
−5y+ 3z= 15
7z= 85
Solving the equations, we find z=85
7. Substitute zback to solve for yand x.
Therefore, y= 4 and x=−5.
Therefore, the solution to the system of equations is x=−5, y= 4, and
z=85
7.
Question 34
Question
Solve the system of linear equations using the Gauss-Jordan elimination method:
2x+ 3y−z= 1
4x+ 2y+ 3z= 4
3x−y+ 2z= 5
Solution
Step 1: Write the augmented matrix for the system of linear equations:
2 3 −1|1
4 2 3 |4
3−1 2 |5
Step 2: Perform row operations to get zeros below the diagonal:
2 3 −1|1
0−4 5 |2
0−10 5 |4
Step 3: Perform row operations to get zeros above the diagonal:
2 3 −1|1
0−4 5 |2
0 0 10 |6
Step 4: Perform row operations to obtain the reduced row-echelon form:
1 0 7
10 |3
5
0 1 −5
8| −1
4
0 0 1 |3
5
33
Step 5: Write the solution back as a system of equations:
x=3
5−7
10z
y=−1
4+5
8z
z=3
5
So, the solution to the system of linear equations is:
x=1
10
y=−1
4
z=3
5
Question 35
Question
Consider the following system of linear equations:
2x+ 3y−z= 7
x−y+ 2z=−1
3x+ 2y+ 4z= 15
Find the solution to the system of equations, or determine if no solution exists.
Solution
Step 1: Write the system of equations in matrix form, Ax=b:
2 3 −1
1−1 2
3 2 4
x
y
z
=
7
−1
15
Step 2: Use Gaussian elimination to row reduce the augmented matrix [A|b].
2 3 −1 7
1−1 2 −1
3 2 4 15
Step 3: Perform row operations to obtain row-echelon form.
1−1 2 −1
0 2.5−2.5 6.5
0 5.5−7 16
34
Step 4: Continue row operations to obtain reduced row-echelon form.
1−1 2 −1
0 1 −1 2.6
0 0 −2 3
Step 5: Back substitute to find the values of x,y, and z: From the last row:
−2z= 3 =⇒z=−3
2
Substitute z=−3
2into the second row: y+3
2= 2.6 =⇒y= 2.6−3
2=5.2
2
Substitute y=5.2
2and z=−3
2into the first row: x−5.2
2−3 = −1 =⇒x=
−1 + 5.2
2+ 3 = 1.8
2
Hence, the solution to the system of equations is: x=9
5,y=26
10 ,z=−3
2.
35
