MATH 112 - TECHNICAL
MATHEMATICS - Applications to
systems of linear equations
Question Bank - Set 4
Liberty University
Question 1
Question
Solve the following system of linear equations:
2x−3y= 5
3x+ 4y= 2
Solution
Step 1: Start by multiplying the first equation by 3 and the second equation
by 2 to make the coefficients of the variable xin both equations the same:
6x−9y= 15
6x+ 8y= 4
Step 2: Subtract the second equation from the first equation to eliminate
x:
0−17y= 11
Step 3: Solve for y:
y=−11
17
Step 4: Substitute the value of yback into one of the original equations to
solve for x. Let’s use the first equation:
2x−3−11
17= 5
Step 5: Simplify and solve for x:
2x+33
17 = 5
2x=85
17
x=85
34
Step 6: The solution to the system of equations is x=85
34 and y=−11
17 .
Question 2
Question
Solve the system of linear equations:
2x−3y+z= 1
x+ 4y−z=−2
3x−2y+ 2z= 5
Solution
Step 1: We can start by writing the system of equations as an augmented matrix:
2−3 1 |1
1 4 −1| −2
3−2 2 |5
Step 2: Perform row operations to simplify the matrix. First, let’s use row
1 to eliminate the first element in rows 2 and 3:
2−3 1 |1
0 11 −3| −4
0 7 −1|2
Step 3: Next, let’s use row 2 to eliminate the second element in row 3:
2−3 1 |1
0 11 −3| −4
0 0 4 |10
Step 4: Now, we can back-substitute to find the values of x,y, and z. From
the last row, we have 4z= 10 ⇒z=10
4=5
2.
Step 5: Substitute z=5
2back into the second row to get 11y−3(5
2) = −4⇒
11y−15
2=−4⇒11y=7
2⇒y=7
22 .
Step 6: Finally, substitute the values of yand zback into the first row to
find x: 2x−3( 7
22 ) + 5
2= 1 ⇒2x−21
22 +5
2= 1 ⇒2x=17
22 ⇒x=17
44 .
Therefore, the solution to the system of equations is x=17
44 ,y=7
22 , and
z=5
2.
2
Question 3
Question
Solve the following system of linear equations:
2x−y+ 3z= 4
3x+ 2y−z= 5
x+ 4y+ 2z= 3
Solution
Step 1: Write the system of equations in matrix form.
2−1 3
3 2 −1
1 4 2
x
y
z
=
4
5
3
Step 2: Row reduce the augmented matrix using elementary row operations.
2−134
3 2 −1 5
1 4 2 3
R2−3
2R1
R3−1
2R1
2−1 3 4
07
2−11
2−1
09
2
1
21
R2↔R3
2−1 3 4
09
2
1
21
07
2−11
2−1
R2−9
14 R3
2−1 3 4
0 1 −1
7
2
7
07
2−11
2−1
R1 + R2
R3−7
2R2
2 0 2 6
0 1 −1
7
2
7
0 0 −9
2−2
R3← −2
9R3
2 0 2 6
0 1 −1
7
2
7
0 0 1 4
9
R1−2R3
R2 + 1
7R3
20010
9
010 6
9
001 4
9
Step 3: The row-reduced echelon form shows the solution x=10
9,y=6
9,
and z=4
9.
3
Question 4
Question
Solve the following system of linear equations:
2x−3y+z= 4
3x+ 2y−2z=−3
x+ 4z= 7
Solution
Step 1: Begin by writing the system of equations in augmented matrix form:
2−3 1 |4
3 2 −2| −3
1 0 4 |7
Step 2: Perform row operations to simplify the matrix.
1 0 4 |7
0 13 −14 | −25
0−3−3| −3
Step 3: Perform additional row operations to further simplify the matrix.
1 0 4 |7
0 1 −1| −1
0 0 1 |1
Step 4: Solve for the variables using back substitution. From the third row,
z= 1. Substitute z= 1 into the second row to solve for y:y−1 = −1 =⇒
y= 0. Finally, substitute z= 1 and y= 0 into the first row to solve for x:
x+ 0 + 4 = 7 =⇒x= 3.
Therefore, the solution to the system of equations is x= 3, y= 0, z= 1.
Question 5
Question
Solve the following system of linear equations:
2x−3y+ 4z= 5
4x+y−z=−3
x+ 2y−3z= 7
4
Solution
Step 1: Write the system of equations in matrix form. Let Abe the coefficient
matrix, Xbe the column matrix of variables, and Bbe the column matrix of
constants.
A=
2−3 4
4 1 −1
1 2 −3
, X =
x
y
z
, B =
5
−3
7
Step 2: Find the determinant of matrix A, denoted as |A|.
|A|=
2−3 4
4 1 −1
1 2 −3
Using cofactor expansion along the first row:
|A|= 2
1−1
2−3
+3
4−1
1−3
= 2((−3)−(−2))+3((−12)−(−1)) = 2(−1)+3(−11) = −2−33 = −35
Step 3: Find the determinants Dx,Dy, and Dzafter replacing the corre-
sponding column in matrix Awith matrix B.
Dx=
5−3 4
−3 1 −1
7 2 −3
, Dy=
2 5 4
4−3−1
1 7 −3
, Dz=
2−3 5
4 1 −3
1 2 7
Step 4: Solve for x,y, and zusing Cramer’s rule.
x=Dx
|A|, y =Dy
|A|, z =Dz
|A|
Substitute the determinants and the determinant of matrix Ainto the for-
mulas above:
x=−35
−35 = 1, y =−39
−35 =39
35 =3
5, z =23
35
Therefore, the solution to the system of equations is x= 1, y=3
5,z=23
35 .
Question 6
Question
Solve the following system of linear equations using the matrix method:
(2x−3y= 7
4x+ 5y= 1
5
Solution
Step 1: Write the system of equations in matrix form.
2−3
4 5 x
y=7
1
Step 2: Set up the augmented matrix.
2−3|7
4 5 |1
Step 3: Perform row operations to get the matrix into row-echelon form.
1−3
2|7
2
0 11 | −13
Step 4: Perform row operations to further simplify the matrix.
1 0 |2
0 1 | −1
Step 5: Write the solution to the system of equations. The solution is x= 2
and y=−1.
Question 7
Question
Solve the following system of linear equations:
(2x−3y= 5
4x+ 5y= 13
Solution
Step 1: To solve the system of equations, we can use the method of substitution.
Let’s solve the first equation for x:
2x−3y= 5 =⇒2x= 3y+ 5 =⇒x=3
2y+5
2
Step 2: Substitute the expression for xinto the second equation:
43
2y+5
2+ 5y= 13
Step 3: Simplify and solve for y:
6y+ 10 + 5y= 13 =⇒11y+ 10 = 13 =⇒11y= 3 =⇒y=3
11
6
Step 4: Substitute y=3
11 back into x=3
2y+5
2to find x:
x=3
2·3
11 +5
2=9
22 +5
2=19
22
Step 5: Therefore, the solution to the system of equations is x=19
22 and
y=3
11 .
Question 8
Question
Solve the system of linear equations:
4x−3y+ 2z= 7
2x+ 5y−z= 4
x−2y+ 3z= 1
Solution
Step 1: Let’s represent the system of equations as an augmented matrix:
4−3 2 |7
2 5 −1|4
1−2 3 |1
Step 2: Perform row operations to simplify the matrix.
4−3 2 |7
2 5 −1|4
1−2 3 |1
Step 3: We will use row operations to get zeros below the diagonal elements
in order to perform back substitution efficiently.
4−3 2 |7
0 11 −5| −10
0 1 5 | −6
Step 4: Perform row operation to get a leading 1 in the second row.
4−3 2 |7
0 1 −5/11 | −10/11
0 1 5 | −6
Step 5: Perform row operations to simplify the matrix further.
4−3 2 |7
0 1 −5/11 | −10/11
0 0 60/11 | −64/11
7
Step 6: Perform row operation to get a leading 1 in the third row.
4−3 2 |7
0 1 −5/11 | −10/11
0 0 1 | −64/60
Step 7: Perform row operations to simplify the matrix further.
4−3 2 |7
0 1 −5/11 | −10/11
0 0 1 | −16/15
Step 8: Perform back substitution to find the values of x,y, and z. From
the last row, we have z=−16
15 . Now substituting the value of zin the second
row, we get y=−10
11 . Finally, substituting the values of yand zin the first row,
we get x= 1.
Therefore, the solution to the system of linear equations is:
x= 1, y =−10
11, z =−16
15
Question 9
Question
Solve the following system of linear equations using the method of your choice:
(5x−2y= 3
3x+ 4y= 6
Solution
Step 1: Let’s solve the system of linear equations using the substitution method.
Step 2: Solve the first equation for x:
5x−2y= 3 =⇒x=3+2y
5
Step 3: Substitute this expression for xinto the second equation:
33+2y
5+ 4y= 6
Step 4: Simplify the equation:
9+6y
5+ 4y= 6
Step 5: Multiply through by 5 to clear the fraction:
9+6y+ 20y= 30
8
Step 6: Combine like terms:
9 + 26y= 30
Step 7: Subtract 9 from both sides:
26y= 21 =⇒y=21
26
Step 8: Substitute y=21
26 back into x=3+2y
5to find x:
x=3+221
26
5
Step 9: Simplify to find the value of x:
x=3 + 42
26
5=3 + 21
13
5=
39+21
13
5=
60
13
5=60
13 ×1
5=12
13
Therefore, the solution to the system of equations is x=12
13 and y=21
26 .
Question 10
Question
Solve the following system of linear equations using the method of your choice:
2x−y+z= 3
4x+ 4y+ 2z= 10
6x+ 3y+ 4z= 12
Solution
Step 1: Let’s write the system of 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
442
634
x
y
z
=
3
10
12
Step 2: We can solve this system by performing row operations to simplify
the matrix to row-echelon form.
2−1 1 |3
442|10
634|12
9
Step 3: Let’s subtract 2 times the first row from the second row, and 3 times
the first row from the third row.
2−1 1 |3
060|4
061|3
Step 4: Next, subtract the second row from the third row to get a new third
row.
2−1 1 |3
060|4
001| −1
Step 5: Now we can back-substitute to find the values of x,y, and z. From
the last row, z=−1. Substituting z=−1 into the second row, we get 6y= 4,
or y=2
3. Finally, substituting y=2
3and z=−1 into the first row, we get
2x= 3 + (2
3)−1 = 4, so x= 2.
Therefore, the solution to the system of equations is x= 2, y=2
3, and
z=−1.
Question 11
Question
Consider the following system of linear equations:
2x−3y+z= 4
3x+ 2y−2z=−7
x+ 4y−3z= 8
Find the solution to the system of equations using the Gaussian elimination
method.
Solution
Step 1: Write the augmented matrix for the system of equations:
2−3 1 |4
3 2 −2| −7
1 4 −3|8
Step 2: Perform row operations to obtain zeros below the diagonal in the
first column:
2−3 1 |4
013
2−7
2| −29
2
1 4 −3|8
10
Step 3: Perform row operations to obtain zeros above the diagonal in the
third column:
2−3 1 |4
013
2−7
2| −29
2
011
2−5
2| −4
Step 4: Perform row operations to obtain zeros below the diagonal in the
second column:
2−3 1 |4
0 13 −7| −29
0 11 −5| −22
Step 5: Perform row operations to solve for x,y, and z:
2−3 1 |4
0 13 −7| −29
0 0 6 |15
Thus, the system has a unique solution given by x= 2, y=−3, and z= 5.
Question 12
Question
Solve the following system of linear equations:
2x+ 3y−z= 7
x−2y+ 3z=−1
3x−y−2z= 8
Solution
Step 1: Begin by writing the system of equations as an augmented matrix.
2 3 −1 7
1−2 3 −1
3−1−2 8
Step 2: Perform row operations to get the matrix into row-echelon form.
1−2 3 −1
0 1 −7 8
0 5 −11 11
Step 3: Continue row operations to further simplify the matrix.
1 0 −17 15
0 1 −7 8
0 0 18 51
11
Step 4: Finish solving by back substitution. From the third row, we find
18z= 51 ⇒z=51
18 =17
6. Then from the second row, we have y−717
6= 8 ⇒
y= 8 + 119
6=47
6. Finally, from the first row, we get x−17 17
6= 15 ⇒x=
15 + 289
6=119
6.
Therefore, the solution to the system of linear equations is x=119
6,y=47
6,
and z=17
6.
Question 13
Question
Solve the following system of linear equations:
2x−3y+z= 7
3x+ 2y= 5
−x+ 4y−3z=−10
Solution
Step 1: Rewrite the system of equations in augmented matrix form:
2−3 1 7
3 2 0 5
−1 4 −3−10
Step 2: Perform row operations to simplify the augmented matrix.
R2−3
2R1→R2
R3+1
2R1→R3
2−3 1 7
013
2−3
2−1
2
05
2−5
2
12
2
Step 3: Continue simplifying the matrix.
R2×2
13 →R2
R3−5
13 R2→R3
2−3 1 7
0 1 −3
13 −1
13
0 0 −1 9
12
Step 4: Back-substitute to find the values of x,y, and z.
z= 9
y−3
13(9) = −1
13
y=2
13
2x−3( 2
13) + 9 = 7
2x=11
13
x=11
26
Therefore, the solution to the system of linear equations is x=11
26 ,y=2
13 ,
and z= 9.
Question 14
Question
Solve the system of linear equations:
2x+ 3y−z= 7
x−2y+ 2z=−1
3x+ 2y+z= 8
Solution
Step 1: Write the system of equations in matrix form AX =B, where Ais the
coefficient matrix, Xis the matrix of variables, and Bis the matrix of constants.
2 3 −1
1−2 2
3 2 1
x
y
z
=
7
−1
8
Step 2: Use row operations to find the row-echelon form of the augmented
matrix.
2 3 −1|7
1−2 2 | −1
3 2 1 |8
1−2 2 | −1
0 4 −5|8
0 8 −5|11
13
1−2 2 | −1
0 1 −5/4|2
0 0 15/4| −5
Step 3: Convert the row-echelon form back into equations and solve for the
variables.
x−2y+ 2z=−1
y−5
4z= 2
15
4z=−5
Solving the last equation, we find z=−20
15 =−4
3.
Substitute z=−4
3into the second equation to solve for y:y−5
4×(−4
3)=2
y+5
3= 2
y= 2 −5
3=1
3
Substitute z=−4
3and y=1
3back into the first equation to solve for x:
x−2×1
3+ 2 ×(−4
3) = −1
x−2
3−8
3=−1
x=−1 + 10
3=7
3
Therefore, the solution to the system of equations is:
x=7
3
y=1
3
z=−4
3
Question 15
Question
Consider the following system of linear equations:
3x−2y+z= 7
x+ 4y−2z=−6
2x−y+ 5z= 15
Solve the system of equations using the method of your choice.
Solution
Step 1: Write the system of equations as an augmented matrix.
3−2 1 7
1 4 −2−6
2−1 5 15
14
Step 2: Perform row operations to row reduce the augmented matrix.
3−2 1 7
1 4 −2−6
2−1 5 15
−−−−−→
R2−1
3R1
3−2 1 7
011
3−7
3−9
2−1 5 15
Step 3: Continue row operations to row reduce the augmented matrix.
3−2 1 7
011
3−7
3−9
2−1 5 15
−−−−−→
R3−2
3R1
3−2 1 7
011
3−7
3−9
01
3
7
31
Step 4: Perform additional row operations.
3−2 1 7
011
3−7
3−9
01
3
7
31
−−−−−→
R2−1
11 R3
3−2 1 7
0 1 −1−6
01
3
7
31
Step 5: Continue row operations.
3−2 1 7
0 1 −1−6
01
3
7
31
−−−−−→
R3−1
3R2
3−2 1 7
0 1 −1−6
0 0 3 3
Step 6: Solve for the variables. From the row echelon form of the augmented
matrix:
3z= 3
z= 1
Substitute the value of z back into the second row:
y−1 = −6
y=−5
Substitute the values of y and z back into the first row:
3x−2(−5) + 1 = 7
3x+ 10 + 1 = 7
3x=−4
x=−4
3
Therefore, the solution to the system of linear equations is:
x=−4
3, y =−5, z =
15
Question 16
Question
Solve the following system of linear equations:
3x+ 2y−z= 1
2x−y+ 2z= 8
x−y+ 3z= 6
Solution
Step 1: Start by writing the augmented matrix of the system of equations.
3 2 −1 1
2−128
1−136
Step 2: Perform row operations to get zeros below the main diagonal.
3 2 −1 1
0−5
3
8
3
38
3
0−5
3
10
3
17
3
Step 3: Perform row operations to get zeros above the main diagonal.
3 2 −1 1
0−5
3
8
3
38
3
0 0 2 5
Step 4: From the row echelon form, we can write the system of equations:
3x+ 2y−z= 1
−5
3y+8
3z=38
3
2z= 5
Step 5: Solve the system by back substitution:
2z= 5
z=5
2
Step 6: Substitute z=5
2back into the second equation to solve for y:
−5
3y+8
35
2=38
3
−5
3y+20
3=38
3
16
−5
3y=18
3
y=−18
5
Step 7: Finally, substitute the values of yand zback into the first equation
to solve for x:
3x+ 2 −18
5−5
2= 1
3x−36
5−5
2= 1
3x=36
5+5
2+ 1
x=36
15
Therefore, the solution to the system of equations is x=12
5,y=−18
5, and
z=5
2.
Question 17
Question
Consider the following system of linear equations:
3x−2y+ 5z= 4
x+ 4y−2z= 9
2x−3y+ 2z= 1
Solve the system of equations using the method of your choice.
Solution
Step 1: Write the augmented matrix for the system of equations.
3−2 5 4
1 4 −2 9
2−3 2 1
Step 2: Apply row operations to get zeros below the diagonal in the first
column.
3−2 5 4
1 4 −2 9
2−3 2 1
17
Multiply the second row by 3 and subtract the first row from it. Multiply
the third row by 2 and subtract the first row from it.
3−2 5 4
0 14 −17 21
0−7−8−7
Step 3: Continue row operations to get zeros below the diagonal in the
second column.
3−2 5 4
0 14 −17 21
0−7−8−7
Multiply the third row by 14 and add the second row to it.
3−2 5 4
0 14 −17 21
0 0 −173 287
Step 4: Solve for z,y, and x. From the last row:
−173z= 287
z=−287
173 =−287
173 =−287
173
Now, substitute z=−287
173 back into the second row:
14y−17 −287
173= 21
14y+4879
173 = 21
14y= 21 −4879
173 =3423
173
y=3423
173 ×14 =3423
2422
Finally, substitute z=−287
173 and y=3423
2422 back into the first row:
3x−23423
2422+ 5 −287
173= 4
Solve for xto find the complete solution to the system of equations.
18
Question 18
Question
Consider the system of equations:
2x+ 3y−z= 7
x−2y+ 2z=−1
4x+y−3z= 8
Solve the system of equations using the method of substitution.
Solution
Step 1: Solve the first equation for xin terms of yand z.
2x=−3y+z+ 7 =⇒x=−3
2y+1
2z+7
2
Step 2: Substitute the expression for xinto the second and third equations.
−3
2y+1
2z+7
2−2y+ 2z=−1
4−3
2y+1
2z+7
2+y−3z= 8
Step 3: Simplify the equations to create a system of two equations in terms
of yand z.
−7
2y+3
2z=−9
2(1)
−5y+ 5z= 4 (2)
Step 4: Solve the system of equations from Step 3. Multiply equation (1)
by 5 and add it to equation (2):
−35
2y+15
2z=−45
2
−35y+ 35z= 20
Step 5: Solve the resulting equation for y.
35z−35
2z= 20 + 45
2=⇒35
2z=85
2=⇒z= 3
Step 6: Substitute z= 3 back into equation (1) to solve for y.
−7
2y+3
2(3) = −9
2=⇒ − 7
2y+9
2=−9
2=⇒ − 7
2y=−6 =⇒y=12
7
19
Step 7: Substitute the values of yand zback into the expression for xfrom
Step 1 to find x.
x=−3
212
7+1
2(3) + 7
2=−18
7+3
2+49
14 =11
7
Step 8: The solution to the system of equations is x=11
7,y=12
7, and
z= 3.
Question 19
Question
Solve the following system of linear equations using the substitution method:
(2x−y= 3
3x+ 2y= 8
Solution
Step 1: Solve the first equation for yin terms of x.
2x−y= 3
⇒y= 2x−3
Step 2: Substitute 2x−3 for yin the second equation and solve for x.
3x+ 2(2x−3) = 8
3x+ 4x−6=8
7x−6=8
7x= 14
x= 2
Step 3: Substitute x= 2 back into the equation y= 2x−3 to solve for y.
y= 2(2) −3
y= 4 −3
y= 1
Therefore, the solution to the system of equations is x= 2 and y= 1.
Question 20
Question
Solve the following system of linear equations:
20
3x+2y−z= 12
2x−y+2z=−5
x+4y−3z= 7
Solution
Step 1: Let’s write the system of equations in matrix form. We have:
3 2 −1
2−1 2
1 4 −3
x
y
z
=
12
−5
7
Step 2: Now, we will augment the matrix with the constants on the right
side:
3 2 −1 12
2−1 2 −5
1 4 −3 7
Step 3: We will perform row operations to row reduce the augmented matrix
to row-echelon form.
Adding −2
3times Row 1 to Row 2:
3 2 −1 12
0−7
3
4
3−13
1 4 −3 7
Multiplying Row 1 by 1
3:
12
3−1
34
0−7
3
4
3−13
1 4 −3 7
Subtracting Row 1 from Row 3:
12
3−1
34
0−7
3
4
3−13
010
3−8
33
Step 4: Next, we will multiply the second row by −3
7:
12
3−1
34
0 1 −4
7
39
7
010
3−8
33
Step 5: Finally, we will perform row operations to obtain the reduced row-
echelon form:
Subtracting 10
3times Row 2 from Row 3:
21
12
3−1
34
0 1 −4
7
39
7
0 0 0 −5
7
Step 6: The last row indicates the equation 0z=−5
7, which is not possible.
Therefore, the system of equations is inconsistent, and there are no solutions.
Question 21
Question
Solve the following system of linear equations using the method of substitution:
(2x+ 3y= 6
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y:
x−y= 1
x=y+ 1
Step 2: Substitute x=y+ 1 into the first equation:
2(y+ 1) + 3y= 6
Step 3: Simplify the equation and solve for y:
2y+ 2 + 3y= 6
5y+ 2 = 6
5y= 4
y=4
5
Step 4: Substitute y=4
5back into x=y+ 1 to find x:
x=4
5+ 1
x=9
5
Therefore, the solution to the system of equations is:
x=9
5, y =4
5
22
Question 22
Question
Consider the following system of linear equations:
2x−y+z= 4
x+y−z=−2
3x−2y+ 2z= 8
Determine whether the system has a unique solution, infinitely many solu-
tions, or no solution. If the system has a unique solution, find the solution for
x,y, and z.
Solution
Step 1: Write the system of equations as an augmented matrix.
2−1 1 4
1 1 −1−2
3−2 2 8
Step 2: Perform row operations to get the matrix into row-echelon form.
2−1 1 4
03
2−3
2−6
0 0 0 0
Step 3: Determine the rank of the coefficient matrix and the rank of the
augmented matrix.
The rank of the coefficient matrix is 2, while the rank of the augmented
matrix is 2.
Step 4: Analyze the ranks to determine the number of solutions.
Since the rank of the coefficient matrix is equal to the rank of the augmented
matrix and both ranks are less than the number of variables, the system has
infinitely many solutions.
Hence, the system of equations has infinitely many solutions.
Question 23
Question
Solve the following system of linear equations:
(2x−3y= 7
4x+y= 5
23
Solution
Step 1: Multiply the second equation by 3 to eliminate y.
(2x−3y= 7
12x+ 3y= 15
Step 2: Add the two equations together to eliminate y.
14x= 22
Step 3: Solve for x.
x=22
14 =11
7
Step 4: Substitute xback into either of the original equations. We will use
the first equation.
211
7−3y= 7
Step 5: Solve for y.
22
7−3y= 7 −3y= 7 −22
7=49
7−22
7=27
7y=27
7·−1
3=−9
7
Step 6: The solution to the system of equations is x=11
7and y=−9
7.
Question 24
Question
Solve the following system of linear equations:
2x−y+ 3z= 5
x+ 2y−z=−4
3x−2y+ 4z= 8
Solution
Step 1: Rewrite the system of equations in matrix form as AX =B, where Ais
the coefficient matrix, Xis the column matrix of variables, and Bis the column
matrix of constants.
A=
2−1 3
1 2 −1
3−2 4
, X =
x
y
z
, B =
5
−4
8
Step 2: Find the determinant of matrix A, denoted by |A|.
|A|=
2−1 3
1 2 −1
3−2 4
24
Step 3: Calculate the determinant using cofactor expansion along the first
row.
|A|= 2
2−1
−2 4
−(−1)
1−1
3 4
+ 3
1 2
3−2
Step 4: Compute the determinants of the 2x2 matrices.
|A|= 2((2 ·4) −(−1)(−2)) −(−1)((1 ·4) −(−1)(3)) + 3((1 · −2) −(2 ·3))
|A|= 2(8 −2) −(−1)(4 + 3) + 3(−2−6)
|A|= 2(6) + 7 −3(−8)
|A|= 12 + 7 + 24
|A|= 43
Step 5: Since |A| = 0, the system has a unique solution. Next, find the
inverse of matrix A, denoted by A−1.
A−1=1
|A|
A11 A21 A31
A12 A22 A32
A13 A23 A33
Step 6: Calculate the inverse matrix using the formula.
A−1=1
43
837
−3 2 −1
−4−1 2
Step 7: Finally, determine the solution matrix Xby multiplying A−1with
matrix B.
X=A−1·B=1
43
837
−3 2 −1
−4−1 2
5
−4
8
After computing the product, we get X=
x
y
z
=
1
−2
3
. Thus, the solution to
the system of equations is x= 1, y=−2, and z= 3.
Question 25
Question
Solve the following system of linear equations using the method of substitution:
(3x+ 4y= 10
2x−3y= 5
25
Solution
Step 1: Solve the first equation for xin terms of y.
3x+ 4y= 10 =⇒3x= 10 −4y=⇒x=10 −4y
3
Step 2: Substitute x=10−4y
3into the second equation.
210 −4y
3−3y= 5
20 −8y
3−3y= 5
20 −8y−9y
3= 5
20 −17y= 15
Step 3: Solve for y.
−17y=−5 =⇒y=−5
−17 =⇒y=5
17
Step 4: Substitute y=5
17 back into x=10−4y
3to find x.
x=10 −4( 5
17 )
3=10 −20
17
3=
170−20
17
3=
150
17
3=150
51 =50
17
Step 5: Therefore, the solution to the system of equations is x=50
17 and
y=5
17 .
Question 26
Question
Solve the following system of linear equations using the matrix method:
2x−3y+z= 4
3x+ 2y−z= 2
x+ 4y−2z= 7
Solution
Step 1: Write the system of equations in matrix form as AX =B, where Ais
the matrix of coefficients, Xis the matrix of variables, and Bis the matrix of
constants.
A=
2−3 1
3 2 −1
1 4 −2
, X =
x
y
z
, B =
4
2
7
26
Step 2: Find the inverse of matrix A, denoted as A−1, by using the formula
A−1=1
det(A)adj(A), where adj(A) is the adjugate matrix of A.
det(A) =
2−3 1
3 2 −1
1 4 −2
=−1
adj(A) =
10 −5 5
−7 4 1
14 −7−4
A−1=1
−1
10 −5 5
−7 4 1
14 −7−4
=
−10 5 −5
7−4−1
−14 7 4
Step 3: Calculate the solution matrix Xby multiplying A−1with B.
X=A−1·B=
−10 5 −5
7−4−1
−14 7 4
·
4
2
7
=
−3
2
1
Step 4: The solution to the system of equations is x=−3, y= 2, z= 1.
Question 27
Question
Solve the following system of linear equations:
2x−y+ 3z= 4
4x+ 5y−z=−2
3x+ 2y+z= 5
Solution
Step 1: We can rewrite the system of equations as an augmented matrix:
2−1 3 |4
4 5 −1| −2
3 2 1 |5
Step 2: To solve the system of equations, we will use row operations to
reduce the matrix to row-echelon form.
Step 3:
2−1 3 |4
4 5 −1| −2
3 2 1 |5
→
1−1
2
3
2|2
0 6 −4| −10
07
2
−1
2|1
2
27
Step 4:
1−1
2
3
2|2
0 6 −4| −10
07
2
−1
2|1
2
→
1−1
2
3
2|2
0 1 −2
3|5
3
0 0 13
3|−11
3
Step 5:
1−1
2
3
2|2
0 1 −2
3|5
3
0 0 13
3|−11
3
→
1−1
2
3
2|2
0 1 −2
3|5
3
0 0 1 | −1
Step 6: The row-echelon form of the augmented matrix corresponds to the
system:
x−1
2y+3
2z= 2
y−2
3z=5
3
z=−1
Step 7: By back substitution, we find the solution to the system as:
x= 1
y= 1
z=−1
Therefore, the solution to the system of linear equations is x= 1, y= 1,
z=−1.
Question 28
Question
Solve the following system of linear equations using the matrix method:
6x−2y+ 3z= 0
2x+ 4y−z= 11
3x−5y+ 2z=−1
Solution
Step 1: Write the system of equations in matrix form. Let Arepresent the
coefficients of the variables, Xrepresent the variables, and Brepresent the
constants.
6−2 3
2 4 −1
3−5 2
x
y
z
=
0
11
−1
28
Step 2: Find the inverse of matrix A.
det(A) = 6(4∗2+5)−(−2)(2∗2+3)+3(2∗−5−3) = 6(13)−(−2)(7)+3(−11) = 78+14−33 = 59
A−1=1
det(A)
4∗2+5 −(2 ∗2 + 3) 2 ∗ −5−3
−(6 ∗2 + 3) 6 ∗2+3 −6∗ −5−3
2∗(−5) 6 ∗2 6 ∗ −5
A−1=1
59
13 −7−11
−15 15 33
−10 12 −30
Step 3: Multiply A−1by matrix Bto solve for X.
X=A−1B=1
59
13 −7−11
−15 15 33
−10 12 −30
0
11
−1
X=1
59
13 ∗0 + −7∗11 + −11 ∗(−1)
−15 ∗0 + 15 ∗11 + 33 ∗(−1)
−10 ∗0 + 12 ∗11 + −30 ∗(−1)
X=1
59
77
132
162
=
77
59
132
59
162
59
Therefore, the solution to the system of equations is x=77
59 ,y=132
59 , and
z=162
59 .
Question 29
Question
Solve the following system of linear equations:
2x−3y+z= 7
−x+ 4y−2z= 4
3x−2y+z= 1
Solution
Step 1: Write the augmented matrix for the system of equations.
2−3 1 7
−1 4 −2 4
3−2 1 1
29
Step 2: Perform row operations to simplify the augmented matrix.
2−3 1 7
−1 4 −2 4
3−2 1 1
→
1−3
2
1
2
7
2
05
2−3
2
15
2
05
2−1
2−13
2
Step 3: Perform additional row operations to further simplify the augmented
matrix.
1−3
2
1
2
7
2
05
2−3
2
15
2
05
2−1
2−13
2
→
1−3
2
1
2
7
2
0 1 −3
53
0 0 4
5−7
Step 4: Convert the augmented matrix back into a system of equations and
solve for x,y, and z.
x−3
2y+1
2z=7
2
y−3
5z= 3
4
5z=−7
From the third equation, we have z=−35
4. Substituting this back into the
second equation gives y−3
5×(−35
4) = 3, which simplifies to y= 4. Finally,
substituting y= 4 and z=−35
4back into the first equation gives x−3
2×4 +
1
2×(−35
4) = 7
2, which simplifies to x= 5.
Therefore, the solution to the system of equations is x= 5, y= 4, and
z=−35
4.
Question 30
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
x−2y+ 3z=−6
3x+y−2z= 1
Solution
Step 1: Begin by expressing the system in matrix form Ax=b, where Ais
the coefficient matrix, xis the column vector of variables, and bis the column
vector of constants.
2 3 −1
1−2 3
3 1 −2
x
y
z
=
4
−6
1
Step 2: Apply row operations to the augmented matrix [A—b] to get it in
reduced row-echelon form.
2 3 −1 4
1−2 3 −6
3 1 −2 1
30
Step 3: Subtract 3×Row 1 from Row 3, and subtract 1
2×Row 1 from Row 2.
2 3 −1 4
0−7
2
7
2−8
0−8 1 −11
Step 4: Divide Row 2 by −7
2, then add 8
7×Row 2 to Row 3.
2 3 −1 4
0 1 −116
7
0 0 −1
7−69
7
Step 5: Divide Row 3 by −1
7to find the solution:
x=−7, y = 9, z = 9
Therefore, the solution to the system of equations is x=−7, y= 9, and
z= 9.
Question 31
Question
Solve the following system of linear equations:
2x−3y+z= 7
4x+y−z= 0
x+ 2y−2z=−4
Solution
Step 1: Write the system of equations in matrix form as AX =B, where Ais
the coefficient matrix, Xis the variable matrix, and Bis the constant matrix.
A=
2−3 1
4 1 −1
1 2 −2
, X =
x
y
z
, B =
7
0
−4
Step 2: To solve the system, we need to find the inverse of matrix A.
det(A) =
2−3 1
4 1 −1
1 2 −2
= 2(1(−2)−(−1)(2))−(−3)(4−1(2)) = 2(−4+2)−(−3)(2) = −8+6 = −2
Step 3: Calculate the cofactor matrix Cby finding the determinant of each
2x2 minor matrix of A.
C=
1−1 5
−3−1−8
246
31
Step 4: Find the adjugate matrix A∗by transposing matrix C.
A∗=CT=
1−3 2
−1−1 4
5−8 6
Step 5: Compute the inverse of Ausing the formula A−1=1
det(A)A∗.
A−1=1
−2
1−3 2
−1−1 4
5−8 6
=
−1
2
3
2−1
1
2
1
2−2
−5
24−3
Step 6: Multiply A−1by Bto find X.
X=A−1B=
−1
2
3
2−1
1
2
1
2−2
−5
24−3
7
0
−4
=
2
−3
−1
Therefore, the solution to the system of linear equations is x= 2, y=−3,
and z=−1.
Question 32
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 2y−z=−1
3x−y+ 2z= 3
Solution
Step 1: Write the system of equations in augmented matrix form.
2−3 1 4
1 2 −1−1
3−1 2 3
Step 2: Perform row operations to get the matrix in row-echelon form.
1 2 −1−1
0−7 3 6
0−7 5 6
Step 3: Continue row operations to reach reduced row-echelon form.
1 2 −1−1
0 1 −3
7−6
7
0 0 1 0
32
Step 4: Write the system of equations represented by the reduced matrix.
x+ 2y−z=−1
y−3
7z=−6
7
z= 0
Step 5: Solve for z in the third equation, then substitute back to find y and
x.
x+ 2y=−1
y=−3
7·0=0
z= 0
Therefore, the solution to the system of equations is x=−1, y= 0, and
z= 0.
Question 33
Question
Solve the following system of linear equations:
2x+ 3y−z= 4
x−2y+ 3z=−6
3x+y+ 2z= 7
Solution
Step 1: Write the system of equations in matrix form. Let
A=
2 3 −1
1−2 3
3 1 2
,X=
x
y
z
,and B=
4
−6
7
Step 2: Write the augmented matrix [A|B].
[A|B] =
2 3 −1 4
1−2 3 −6
3 1 2 7
Step 3: Perform row operations to get the matrix into row-echelon form.
R2−1
2R1→R2
R3−3
2R1→R3
33
The new augmented matrix is:
2 3 −1 4
0−7
2
7
2−8
0−7
2
7
2−1
Step 4: Perform additional row operations to simplify the matrix further.
R3−R2→R3
The new augmented matrix is:
2 3 −1 4
0−7
2
7
2−8
0 0 0 7
Step 5: Write the system of equations represented by the row-echelon form
of the matrix.
2x+ 3y−z= 4
−7
2y+7
2z=−8
This system simplifies to 0 = 7, which is a contradiction.
Step 6: Since the system simplifies to a false statement, there is no solution
to the system of equations. The system is inconsistent.
Question 34
Question
Solve the following system of linear equations using the Gaussian elimination
method:
4x−2y+ 3z= 7
2x+ 3y−z= 6
−3x+y+ 2z= 4
Solution
Step 1: Write the augmented matrix for the system of equations:
4−2 3 |7
2 3 −1|6
−3 1 2 |4
Step 2: Perform row operations to obtain zeros below and above the leading
1 in the first column:
4−2 3 |7
0 4 −5| −8
0−5 11 |25
34
Step 3: Perform row operations to obtain zeros below and above the leading
1 in the second column:
4−2 3 |7
0 4 −5| −8
0 0 1 |1
Step 4: Perform row operations to obtain zeros above the leading 1 in the
third column:
4−2 0 |5
040| −4
001|1
Step 5: Perform row operations to obtain echelon form:
1−1
20|5
4
010| −1
001|1
Step 6: Back substitute to find the values of x,y, and z:
x=5
4
y=−1
z= 1
Therefore, the solution to the system of linear equations is x=5
4,y=−1,
and z= 1.
Question 35
Question
Solve the following system of linear equations using the matrix method:
3x+ 2y−z= 6
2x−2y+ 4z=−7
x+ 3y+ 2z= 4
Solution
Step 1: Write the system of equations in matrix form AX =B, where
A=
3 2 −1
2−2 4
1 3 2
, X =
x
y
z
, B =
6
−7
4
35
Step 2: Find the inverse of matrix Aby performing row operations on the
augmented matrix [A|I].
3 2 −1 1 0 0
2−2 4 0 1 0
1 3 2 0 0 1
3 2 −1 1 0 0
0−8 6 −210
1 3 2 0 0 1
3 2 −1 1 0 0
0−8 6 −210
0 7 3 −101
3 2 −1 1 0 0
0−8 6 −2 1 0
0 0 51 −17 7 3
3 2 −1 1 0 0
0−8 0 −8 5 6
0 0 1 −17
51
7
51
3
51
3 0 −144
51 −14
51 −3
51
0−8 0 −8 5 6
0 0 1 −17
51
7
51
3
51
100 5
17 −7
17 −1
17
0 1 0 1 −5
8−3
4
001−17
51
7
51
3
51
Step 3: The solution to the system is x=5
17 ,y= 1, and z=−1
3.
36
