MATH 112 - TECHNICAL
MATHEMATICS - Applications to
systems of linear equations
Question Bank - Set 3
Liberty University
Question 1
Question
Solve the following system of linear equations using the matrix method:
2x+ 3y+z= 4
x−2y+ 3z= 1
3x+y−2z= 6
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2 3 1 |4
1−2 3 |1
3 1 −2|6
Step 2: Perform row operations to get the matrix into row-echelon form.
1−2 3 |1
2 3 1 |4
3 1 −2|6
Step 3: Use row operations to further simplify the matrix to reduced row-
echelon form.
1−2 3 |1
0 7 −5|2
0 7 −13 |3
Step 4: Continue reducing to fully row-reduced echelon form.
1−2 3 |1
0 1 −5
7|2
7
0 0 −48
7|11
7
Step 5: Back-substitute to solve for z,y, and x. From the last row, we have
−48
7z=11
7. So z=−11
48 . Substitute zback into the second row gives y=2
7.
Finally, substitute yand zback into the first row gives x= 2.
Therefore, the solutions to the system of linear equations are x= 2, y=2
7,
and z=−11
48 .
Question 2
Question
Solve the following system of linear equations using the method of your choice:
2x+ 4y−z= 7
x−3y+ 2z=−1
3x+ 2y−2z= 4
Solution
Step 1: Rearrange the system of equations in matrix form.
2 4 −1
1−3 2
3 2 −2
x
y
z
=
7
−1
4
Step 2: Use Gaussian elimination to transform the augmented matrix to
row-echelon form.
2 4 −1|7
1−3 2 | −1
3 2 −2|4
→
1−3/2 1/2|2
0 5/2 3/2|3
0 11/2−7/2| −2
Step 3: Continue using Gaussian elimination to reach row-echelon form.
1−3/2 1/2|2
0 1 3/5|6/5
0 0 −32/5| −32/5
Step 4: Back substitute and solve for the variables.
From the last row, we get −32/5z=−32/5, so z= 1.
Substituting z= 1 into the second row, we get y+ 3/5 = 6/5, so y= 1.
Substituting y= 1 and z= 1 back into the first row, we get x−3/2+1/2 = 2,
so x= 2.
Therefore, the solution to the system of equations is x= 2, y= 1, and z= 1.
2
Question 3
Question
Solve the following system of linear equations:
2x−3y+ 5z= 4 (1)
4x+ 2y−z=−7 (2)
3x−4y+ 2z= 10 (3)
Solution
Step 1: Begin by rewriting 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.
2−3 5
4 2 −1
3−4 2
x
y
z
=
4
−7
10
Step 2: Use Gaussian elimination to row reduce the augmented matrix [A|b]
to row-echelon form:
2−3 5 |4
4 2 −1| −7
3−4 2 |10
Divide row 1 by 2:
1−3
2
5
2|2
4 2 −1| −7
3−4 2 |10
Subtract 4 times row 1 from row 2 and 3 times row 1 from row 3:
1−3
2
5
2|2
0 8 −11 | −15
0−1
2−1
2|4
Step 3: Continue row operations to further simplify the matrix to reduced
row-echelon form: Divide row 2 by 8 and row 3 by -0.5:
1−3
2
5
2|2
0 1 −11
8| −15
8
0 1 1 | −8
Subtract row 2 from row 3:
1−3
2
5
2|2
0 1 −11
8| −15
8
0 0 19
8| −97
8
3
Step 4: Solve for zusing the last row:
19
8z=−97
8=⇒z=−5
Step 5: Substitute z=−5 into the second row to solve for y:
y−11
8(−5) = −15
8=⇒y+55
8=−15
8=⇒y=−5
Step 6: Substitute z=−5 and y=−5 into the first row to solve for x:
x−3
2(−5) + 5
2(−5) = 2 =⇒x+15
2−25
2= 2 =⇒x−5 = 2 =⇒x= 7
Therefore, the solution to the given system of equations is x= 7, y=−5,
and z=−5.
Question 4
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
4x−y+ 3z=−4
x+ 2y−2z= 2
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
4−1 3
1 2 −2
, X =
x
y
z
, B =
1
−4
2
Step 2: Find the inverse of matrix Aby calculating A−1.
det(A) =
2 3 −1
4−1 3
1 2 −2
= 18
A−1=1
det(A)
A11 A21 A31
A12 A22 A32
A13 A23 A33
A−1=1
18
−8 9 1
−14 6 −2
−5 4 −2
4
Step 3: Solve for Xby multiplying X=A−1B.
X=
−8 9 1
−14 6 −2
−5 4 −2
1
−4
2
=
−17
−18
3
Therefore, the solution to the system of equations is x=−17, y=−18, and
z= 3.
Question 5
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 2y−z=−1
3x−y+ 2z= 7
Solution
Step 1: Write the system of equations in matrix form as Ax=bwhere
A=
2−3 1
1 2 −1
3−1 2
,x=
x
y
z
,and b=
4
−1
7
Step 2: We want to find xsuch that Ax=b. Let’s find the inverse of
matrix Afirst.
Find det(A) = det
2−3 1
1 2 −1
3−1 2
= 29
Step 3: Find the adjugate matrix A∗of matrix A.
A∗=
3 7 5
−163
−5−8 4
T
=
3−1−5
7 6 −8
5 3 4
Step 4: Calculate the inverse of matrix Ausing the formula: A−1=
1
det(A)A∗.
A−1=1
29
3−1−5
7 6 −8
5 3 4
=
3
29 −1
29 −5
29
7
29
6
29 −8
29
5
29
3
29
4
29
Step 5: Multiply both sides of the equation Ax=bby A−1.
A−1(Ax) = A−1b
5
x=
3
29 −1
29 −5
29
7
29
6
29 −8
29
5
29
3
29
4
29
4
−1
7
Step 6: Perform matrix multiplication to find the values of x,y, and z.
This will give the solution to the system of equations.
Question 6
Question
Solve the following system of linear equations:
3x+ 2y+z= 7
x−y+ 2z= 3
2x+ 3y−z= 8
Solution
Step 1: Write the system of equations in matrix form as AX =B, where A
is the matrix of coefficients, Xis the column matrix of variables, and Bis the
column matrix of constants.
3 2 1
1−1 2
2 3 −1
x
y
z
=
7
3
8
Step 2: Find the inverse of matrix Ato solve for X.
A−1=1
det(A)adj(A)
First, calculate the determinant of matrix A:
det(A) = 3(−1)(−1) + 2(2)(2) + 1(1)(3) −1(−1)(2) −2(1)(3) −3(2)(−1) = 20
Step 3: Calculate the adjugate of matrix A.
adj(A) =
(−1)1+1M11 (−1)1+2M12 (−1)1+3M13
(−1)2+1M21 (−1)2+2M22 (−1)2+3M23
(−1)3+1M31 (−1)3+2M32 (−1)3+3M33
where Mij are the determinants of the (i, j) minors of matrix A.
Step 4: Calculate the inverse of matrix Ausing the formula A−1=1
det(A)adj(A).
A−1=1
20
6 1 5
5 3 1
−4−2 2
6
Step 5: Multiply both sides of the equation AX =Bby A−1to solve for X.
X=A−1B=1
20
6 1 5
5 3 1
−4−2 2
7
3
8
Step 6: Compute the product A−1Bto find the values of x,y, and z.
Question 7
Question
Find the solution to the system of linear equations:
3x−2y+z= 5
2x+y−3z=−1
x−4y+ 2z= 7
Solution
Step 1: Write the system of linear equations in matrix form as AX =B, where
Ais the coefficient matrix, Xis the variable matrix, and Bis the constant
matrix.
A=
3−2 1
2 1 −3
1−4 2
, X =
x
y
z
, B =
5
−1
7
Step 2: Find the inverse of matrix Aby calculating det(A) and A−1.
det(A) = 3(1(2) −(−3)(−4)) −(−2)(2(2) −(−3)(1)) + 1(2(−4) −1(−2)) = 47
A−1=1
det(A)
1(2) −(−3)(−4) −2(2) −(−3)(1) −2(−4) −1(−2)
−2(2) −(−3)(1) 3(2) −1(1) 1(−2) −3(−4)
1(−4) −2(−2) 2(−4) −1(1) 2(1) −2(2)
A−1=1
47
14 −7 6
−1 5 10
0−9−2
Step 3: Calculate the solution matrix X=A−1B.
X=1
47
14 −7 6
−1 5 10
0−9−2
5
−1
7
=1
47
(14 ·5) + (−7· −1) + (6 ·7)
(−1·5) + (5 · −1) + (10 ·7)
(0 ·5) + (−9· −1) + (−2·7)
7
X=1
47
70 + 7 + 42
−5−5 + 70
0+9−14
=1
47
119
60
23
Therefore, the solution to the system of linear equations is x=119
47 ,y=60
47 ,
and z=23
47 .
Question 8
Question
Solve the following system of linear equations:
2x−y+z= 5
x+ 3y−2z=−2
3x+y+ 4z= 10
Solution
Step 1: Rewrite the system of equations in matrix form:
2−1 1
1 3 −2
3 1 4
x
y
z
=
5
−2
10
Step 2: Use elementary row operations to row reduce the augmented matrix.
2−1 1 |5
1 3 −2| −2
3 1 4 |10
Step 3: Perform row operations to row reduce the matrix further.
1 3 −2| −2
0−5 5 |4
0−8 10 |16
Step 4: Continue row operations to simplify the matrix.
1 3 −2| −2
0 1 −1| −4/5
0 0 3 |12/5
Step 5: Simplify the equations back to variable form. From the third row,
3z=12
5⇒z=4
5. Substitute z=4
5back into the second row to solve for y:
y−4
5=−4
5⇒y= 0. Finally, substitute y= 0 and z=4
5into the first row to
solve for x: 2x−0 + 4
5= 5 ⇒x=23
10 .
Step 6: Therefore, the solution to the system of equations is x=23
10 , y =
0, z =4
5.
8
Question 9
Question
Solve the following system of linear equations:
2x−3y+z= 6
x+ 2y−2z=−8
3x−y+ 4z= 9
Solution
Step 1: Write the system of equations in matrix form:
2−3 1
1 2 −2
3−1 4
x
y
z
=
6
−8
9
Step 2: Row-reduce the augmented matrix:
2−3 1 6
1 2 −2−8
3−1 4 9
→
1 2 −2−8
0−1 5 22
0−7 10 33
→
1 2 −2−8
0 1 −5−22
0 0 45 121
Step 3: Perform back substitution to solve for z:
45z= 121 ⇒z=121
45 =121
45
Step 4: Substitute z=121
45 back into the second row to solve for y:
y−5121
45 =−22 ⇒y=−22 + 5 121
45 =77
9
Step 5: Substitute z=121
45 and y=77
9back into the first row to solve for
x:
x+ 2 77
9−2121
45 =−8⇒x=−8−154
9+242
45 =−111
5
Step 6: Therefore, the solution to the system of equations is x=−111
5,
y=77
9, and z=121
45 .
Question 10
Question
Find the values of xand ythat satisfy the following system of linear equations:
(3x−2y= 7
2x+ 4y= 1
9
Solution
To solve this system, we can use the method of substitution or the method of
elimination. Here, we will use the method of elimination.
Step 1: Multiply the first equation by 2 and the second equation by 3 to
create opposite coefficients for yin each equation.
(6x−4y= 14
6x+ 12y= 3
Step 2: Subtract the two equations to eliminate y.
−16y= 11
Step 3: Solve for y.
y=−11
16
Step 4: Substitute yback into one of the original equations to solve for x.
3x−2−11
16= 7
Step 5: Simplify and solve for x.
3x+22
16 = 7
3x= 7 −22
16
3x=82
16
x=82
48 =41
24
Step 6: Therefore, the solution to the system of equations is x=41
24 and
y=−11
16 .
Question 11
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
5x−y+ 3z= 2
2x+ 3y+z= 1
10
Solution
Step 1: We can write the system of equations in matrix form as Ax=bwhere
Ais the coefficient matrix, xis the column vector of variables, and bis the
column vector of constants.
3 2 −1
5−1 3
2 3 1
x
y
z
=
4
2
1
Step 2: To solve for x, we find the inverse of matrix A:
A−1=1
det(A)adj(A)
where det(A) is the determinant of Aand adj(A) is the adjugate of A.
Step 3: First, calculate the determinant of matrix A:
det(A) =
3 2 −1
5−1 3
2 3 1
=−59
Step 4: Next, find the matrix of cofactors, which is the transpose of the
matrix of minors:
cof(A) =
−8−11 13
1 5 −8
11 13 −9
Step 5: Calculate the adjugate by taking the transpose of the matrix of
cofactors:
adj(A) =
−8 1 11
−11 5 13
13 −8−9
Step 6: Now, find the inverse of matrix A:
A−1=1
−59
−8 1 11
−11 5 13
13 −8−9
=
8
59 −1
59 −11
59
11
59 −5
59 −13
59
−13
59
8
59
9
59
Step 7: Finally, solve for xby multiplying the inverse of matrix Awith b:
x
y
z
=
8
59 −1
59 −11
59
11
59 −5
59 −13
59
−13
59
8
59
9
59
4
2
1
After performing the matrix multiplication, we find the solution to be x=
−1, y = 2, z = 3.
11
Question 12
Question
Solve the following system of linear equations:
2x−3y+z= 7
4x+ 2y−3z=−2
x+ 5y−2z= 9
Solution
Step 1: Write the system of equations in augmented matrix form:
2−3 1 |7
4 2 −3| −2
1 5 −2|9
Step 2: Perform row operations to get leading 1s in the first column:
1−3
2
1
2|7
2
0 7 −4| −9
013
2−5
2|5
2
Step 3: Use row operations to get 0s below the leading 1 in the first column:
1−3
2
1
2|7
2
0 7 −4| −9
0 0 1 |3
Step 4: Continue row operations to get the reduced row-echelon form:
100|2
010| −1
001|3
Step 5: The solution to the system of equations is x= 2, y =−1, z = 3.
Question 13
Question
Solve the following system of linear equations using the method of elimination:
3x+ 2y−z= 7
2x−y+ 3z=−1
x+ 4y+ 2z= 18
12
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate xin the first two equations.
6x+ 4y−2z= 14
6x−3y+ 9z=−3
x+ 4y+ 2z= 18
Step 2: Subtract the second equation from the first equation to eliminate x.
6x+ 4y−2z= 14
−7y−11z=−17
x+ 4y+ 2z= 18
Step 3: Multiply the second equation by 4 and add it to the third equation
to eliminate y.
6x+ 4y−2z= 14
−7y−11z=−17
17z= 26
Step 4: Solve the third equation for z.
z=26
17 =2
17
Step 5: Substitute z=2
17 back into the second equation to solve for y.
−7y−11 2
17=−17
−7y−22
17 =−17
−7y=−17 + 22
17
−7y=−243
17
y=243
119
Step 6: Substitute y=243
119 and z=2
17 back into the first equation to solve
for x.
6x+ 4 243
119−22
17= 14
6x+972
119 −4
17 = 14
13
6x= 14 −972
119 +4
17
6x=260
119
x=130
119
Therefore, the solution to the system of linear equations is x=130
119 ,y=243
119 ,
and z=2
17 .
Question 14
Question
Solve the following system of linear equations:
2x+ 3y+z= 15
4x−y+ 2z=−1
x−2y+ 3z= 10
Solution
Step 1: Begin by writing the system of equations in matrix form:
231
4−1 2
1−2 3
x
y
z
=
15
−1
10
Step 2: To solve for x,y, and z, we will perform row operations to put the
system into row-echelon form.
Step 3: Replace R2with R2−2R1and R3with R3−1
2R1:
2 3 1
0−7 0
0−3.5 2.5
x
y
z
=
15
−31
2.5
Step 4: Next, replace R3with R3+ 0.5R2:
2 3 1
0−7 0
002.5
x
y
z
=
15
−31
−15.75
Step 5: Now, back substitute to find the values of x,y, and z:
2z=−15.75
−7y=−31
2x+ 3y+z= 15
Solving, we find z=−7.875, y= 4.429, and x= 3.718.
Therefore, the solution to the system of linear equations is x= 3.718, y=
4.429, and z=−7.875.
14
Question 15
Question
Solve the following system of linear equations:
2x−3y+z= 8
x+ 4y−2z=−2
3x−2y+ 5z= 10
Solution
Step 1: Start by writing the system of equations in matrix form:
2−3 1
1 4 −2
3−2 5
x
y
z
=
8
−2
10
Step 2: Use row operations to simplify the matrix:
2−3 1 8
1 4 −2−2
3−2 5 10
Step 3: Subtract 0.5×Row 1 from Row 2 and 1.5×Row 1 from Row 3:
2−3 1 8
0 5.5−2.5−6
0 2.5 3.5−2
Step 4: Divide Row 2 by 5.5 and subtract 2.5×Row 2 from Row 3:
2−3 1 8
0 1 −0.4545 −1.0909
0 0 4.3636 0.2727
Step 5: Divide Row 3 by 4.3636 and perform back substitution to find x,y,
and z:
2−3 1 8
0 1 −0.4545 −1.0909
0 0 1 0.0625
So, z= 0.0625.
Step 6: Substitute zback into Row 2 to find y:
y−0.4545(0.0625) = −1.0909
Solving for y, we get y=−1.
15
Step 7: Substitute yand zback into Row 1 to find x:
2x−3(−1) + 0.0625 = 8
Solving for x, we get x= 3.9375.
Therefore, the solution to the system of linear equations is:
x= 3.9375, y =−1, z = 0.0625
Question 16
Question
Consider the following system of linear equations:
2x−3y+z= 5
5x+ 2y−3z= 7
x−4y+ 2z= 1
Determine whether the system has a unique solution, no solution, or infinitely
many solutions. If a unique solution exists, solve the system.
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
5 2 −3
1−4 2
x
y
z
=
5
7
1
Step 2: Calculate the determinant of the coefficient matrix.
det
2−3 1
5 2 −3
1−4 2
= 2(2(2)−(−3)(−4))−(−3)(5(2)−1(−4))+1(5(−4)−2(1))
= 2(4−12)−(−15 −(−4))+(20−2) = 2(−8)−(−11)+18 = −16 +11+18 = 13
Step 3: Since the determinant is non-zero (13 = 0), the system has a unique
solution.
Step 4: Use Cramer’s Rule to find the values of x, y, and z:
x=
det
5−3 1
7 2 −3
1−4 2
det
2−3 1
5 2 −3
1−4 2
=13
13 = 1
16
y=
det
2 5 1
5 7 −3
1 1 2
det
2−3 1
5 2 −3
1−4 2
=17
13
z=
det
2−3 5
527
1−4 1
det
2−3 1
5 2 −3
1−4 2
=23
13
Therefore, the system of equations has a unique solution: x= 1, y=17
13 ,
z=23
13 .
Question 17
Question
Solve the system of linear equations using the method of substitution:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute the expression for yinto the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Solve for x.
14x−15 = 7
14x= 22
x=22
14
x=11
7
17
Step 4: Substitute the value of xback into the expression for yto find y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is:
(x=11
7
y=9
7
Question 18
Question
Solve the following system of linear equations:
3x+ 4y−z= 2
x−2y+ 3z=−4
2x+y= 1
Solution
Step 1: Write the system of equations as an augmented matrix:
3 4 −1 2
1−2 3 −4
2 1 0 1
Step 2: Perform row operations to get the matrix into row-echelon form.
1−2
301
3
010
3−1−10
3
05
311
3
Step 3: Continue row operations to get the matrix into reduced row-echelon
form.
1 0 2
5−1
5
0 1 −3
5−1
0 0 0 0
Step 4: Write the system of equations represented by the reduced matrix:
x+2
5z=−1
5
y−3
5z=−1
0 = 0
18
Step 5: Solve the system of equations. First, express yin terms of z:
y=3
5z−1
Step 6: Substitute yinto the equation for x:
x+2
5z=−1
5
x=−2
5z−1
5
Step 7: Therefore, the solution to the system of linear equations is:
x=−2
5z−1
5, y =3
5z−1, z =z
Question 19
Question
Solve the following system of linear equations:
2x−3y+z= 1
x+ 2y−z= 4
3x−y+ 2z= 5
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 1
1 2 −1 4
3−1 2 5
Step 2: Perform row operations to simplify the matrix.
2−3 1 1
1 2 −1 4
3−1 2 5
→
1 2 −1 4
2−3 1 1
3−1 2 5
Step 3: Perform row operations to further simplify the matrix.
1 2 −1 4
2−3 1 1
3−1 2 5
→
1 2 −1 4
0−7 3 −7
0−7 5 −7
Step 4: Continue performing row operations to row reduce the matrix.
1 2 −1 4
0−7 3 −7
0−7 5 −7
→
1 2 −1 4
0 1 −3/7 1
0−7 5 −7
→
1 2 −1 4
0 1 −3/7 1
0 0 26/7 0
19
Step 5: Convert the row reduced matrix back into a system of equations.
x+ 2y−z= 4
y−3
7z= 1
26
7z= 0
Step 6: Solve for the variables. From the third equation, we have z= 0.
Substituting this into the second equation, we get y−3
7(0) = 1, which simplifies
to y= 1. Finally, substituting z= 0 and y= 1 into the first equation, we get
x+ 2(1) −0 = 4, which simplifies to x= 2.
Therefore, the solution to the system of equations is x= 2, y= 1, and z= 0.
Question 20
Question
Solve the following system of linear equations:
2x−y+ 3z= 7
x+ 2y−z= 1
3x+y+ 2z= 11
Solution
Step 1: Rewrite the system of equations as an augmented matrix.
2−1 3 7
1 2 −1 1
3 1 2 11
Step 2: Perform row operations to obtain echelon form.
1 2 −1 1
0−5 4 6
0−5 5 8
Step 3: Continue row operations to reach reduced row-echelon form.
1 0 0 3
0 1 0 2
0 0 1 1
Step 4: Write the solution for the system of equations. The solution is x= 3,
y= 2, and z= 1.
20
Question 21
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+ 2y−z=−1
3x−y= 6
Solution
Let’s solve the system of equations using the method of substitution.
Step 1: Let’s rewrite the system of equations:
2x−3y+z= 7 (1)
x+ 2y−z=−1 (2)
3x−y= 6 (3)
Step 2: From equation (3), express yin terms of x:
y= 3x−6
Step 3: Substitute y= 3x−6 into equation (2):
x+ 2(3x−6) −z=−1
x+ 6x−12 −z=−1
7x−z= 11 (4)
Step 4: Substitute y= 3x−6 into equation (1):
2x−3(3x−6) + z= 7
2x−9x+ 18 + z= 7
−7x+z=−11 (5)
Step 5: Now we have the system of equations:
7x−z= 11 (4)
−7x+z=−11 (5)
Step 6: Add equations (4) and (5) to eliminate z:
0=0
Since 0 = 0 is a true statement, the system of equations has infinitely many
solutions.
21
Question 22
Question
Solve the following system of linear equations:
2x−3y+z= 5
4x+y+ 2z= 4
3x+ 2y−z= 1
Solution
Step 1: Rewrite the system of equations in augmented matrix form.
2−315
4 1 2 4
3 2 −1 1
Step 2: Perform row operations to get zeros below the main diagonal.
2−3 1 5
0 7 0 −6
0 11 −4−14
Step 3: Perform row operations to get zeros above the main diagonal.
2−3 1 5
0 7 0 −6
0 0 −4 2
Step 4: Perform row operations to get a leading 1 in each row.
1−3
2
1
2
5
2
0 1 0 −6
7
0 0 1 −1
2
Step 5: Read off the solutions from the row-echelon form. The solution to
the system of equations is x=5
2,y=−6
7, and z=−1
2.
Question 23
Question
Find all values of ksuch that the system of linear equations is consistent:
2x−y+z= 3
4x+ky + 6z= 10
3x+ 2y+ 3z=k
22
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−1 1 |3
4k6|10
323|k
Step 2: Perform row operations to simplify the matrix. First, replace the
second row with the result of subtracting twice the first row from it:
2−1 1 |3
0k+ 2 4 |4
323|k
Step 3: Next, replace the third row with the result of subtracting 3
2times
the first row from it:
2−1 1 |3
0k+ 2 4 |4
07
2
1
2|k−9
2
Step 4: Divide the third row by 7 to simplify the matrix further:
2−1 1 |3
0k+ 2 4 |4
0 1 1
2|k
7−9
14
Step 5: To ensure the system is consistent, the last row cannot have all zeros
on the left side while a non-zero value on the right. For this to happen, k+ 2
must be non-zero. Thus, we set up the equation k+ 2 = 0 and solve for k:
k+ 2 = 0
k=−2
Therefore, the system of linear equations is consistent for all values of k
except for k=−2.
Question 24
Question
Solve the system of linear equations:
2x−3y+ 4z= 7
x+y−z= 1
3x+ 2y−2z= 6
23
Solution
Step 1: Rewrite the system of equations in matrix form:
2−3 4
1 1 −1
3 2 −2
x
y
z
=
7
1
6
Step 2: Use row operations to row-reduce the augmented matrix:
2−347
1 1 −1 1
3 2 −2 6
R2=R2−1
2R1R3=R3−3
2R1
2−3 4 7
07
2−3−3
013
2−8−3
R3=R3−13
7R2
2−3 4 7
07
2−3−3
0 0 1 0
Step 3: Solve for z:7
2z−3 = −3
7
2z= 0
z= 0
Step 4: Substitute z= 0 back into the second row to solve for y:
7
2y−3 = −3
7
2y= 0
y= 0
Step 5: Substitute y= 0 and z= 0 back into the first row to solve for x:
2x−3(0) + 4(0) = 7
2x= 7
x=7
2
Therefore, the solution to the system of equations is x=7
2,y= 0, and
z= 0.
24
Question 25
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−y+z= 3
2x+ 3y+z= 0
Solution
Step 1: Write the system of linear equations in matrix form.
3 2 −1
1−1 1
2 3 1
x
y
z
=
4
3
0
Step 2: Use row operations to solve the system of equations by reducing
the augmented matrix to row-echelon form.
3 2 −1|4
1−1 1 |3
2 3 1 |0
Step 3: Subtract 2×Row 2 from Row 1:
1 4 −3| −2
1−1 1 |3
2 3 1 |0
Step 4: Subtract 2×Row 2 from Row 3:
1 4 −3| −2
1−1 1 |3
0 5 −1| −6
Step 5: Subtract 1/5×Row 3 from Row 2:
1 4 −3| −2
1 0 2 |4
0 1 −1/5|6/5
Step 6: Subtract 4×Row 3 from Row 1:
1 0 7/5| −22/5
1 0 2 |4
0 1 −1/5|6/5
25
Step 7: Simplify the augmented matrix and write the system of equations
back out: x+7
5z=−22
5
y−1
5z=6
5
Step 8: Solve for yin terms of zfrom the second equation:
y=1
5z+6
5
Step 9: Substitute yinto the first equation to solve for x:
x+7
5z=−22
5
x=−7
5z−22
5
Step 10: The general solution is:
x=−7
5z−22
5, y =1
5z+6
5, z is free
Question 26
Question
Consider the following system of linear equations:
2x+ 3y+z= 6
x−2y+ 2z= 5
3x+y−z= 1
Determine if the system has a unique solution, infinitely many solutions, or no
solution. If a unique solution exists, solve the system.
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 1
1−2 2
3 1 −1
x
y
z
=
6
5
1
Step 2: Next, find the determinant of the coefficient matrix to determine the
type of solution the system has. The determinant Dis given by:
D=
2 3 1
1−2 2
3 1 −1
26
Step 3: Calculate the determinant using cofactor expansion along the first
row:
D= 2
−2 2
1−1
−3
1 2
3−1
+ 1
1−2
3 1
Step 4: Simplify the determinants and calculate D:
D= 2((−2)(−1) −(1)(2)) −3((1)(−1) −(3)(2)) + 1((1)(1) −(3)(−2))
D= 2(2 −2) −3(−1−6) + 1(1 + 6)
D= 0 + 21 + 7
D= 28
Step 5: Since D= 0, the system has a unique solution.
Step 6: Find the solution by calculating x,y, and zusing Cramer’s Rule:
x=Dx
D, y =Dy
D, z =Dz
D
where Dx,Dy, and Dzare the determinants obtained by replacing the x-column,
y-column, and z-column of the coefficient matrix with the constant vector.
Step 7: Calculate Dx,Dy, and Dz:
Dx=
6 3 1
5−2 2
1 1 −1
=−24
Dy=
2 6 1
1 5 2
3 1 −1
= 54
Dz=
236
1−2 5
311
=−32
Step 8: Calculate x,y, and z:
x=Dx
D=−24
28 =−6
7
y=Dy
D=54
28 =27
14
z=Dz
D=−32
28 =−8
7
Therefore, the system has a unique solution: x=−6
7,y=27
14 ,z=−8
7.
27
Question 27
Question
Solve the following system of linear equations:
2x−y+z= 3
4x+y−2z=−1
x+ 2y+z= 4
Solution
Step 1: Write the system of equations in matrix form:
2−1 1
4 1 −2
1 2 1
x
y
z
=
3
−1
4
Step 2: Use Gaussian elimination to row-reduce the augmented matrix:
2−1 1 |3
4 1 −2| −1
1 2 1 |4
R2−2R1
R3−1
2R1→
2−1 1 |3
0 3 −4| −7
05
2
1
2|5
2
R3−5
6R2→
2−1 1 |3
0 3 −4| −7
0 0 1 |3
Step 3: Back-substitute to find the values of x,y, and z: From the third
row, we have z= 3. Substitute z= 3 into the second row to find y:
3y−4(3) = −7
3y= 5
y=5
3
Substitute y=5
3and z= 3 into the first row to solve for x:
2x−5
3+ 3 = 3
2x=2
3
x=1
3
Therefore, the solution to the system of equations is x=1
3,y=5
3, and
z= 3.
28
Question 28
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
4x−y+ 2z=−2
x+ 3y−3z= 0
Solution
Step 1: Write the system of equations in matrix form.
2 3 −1
4−1 2
1 3 −3
x
y
z
=
1
−2
0
Step 2: Use row reduction to find the reduced row echelon form of the
augmented matrix.
2 3 −1|1
4−1 2 | −2
1 3 −3|0
1 0 1 |1
0 1 −2|1
0 0 0 |0
Step 3: Write the system of equations corresponding to the reduced row
echelon form.
x+z= 1
y−2z= 1
0=0
Step 4: Solve the system of equations. From the last equation, we can see
that z is a free variable. Let z = t where t is a parameter.
Step 5: Express x and y in terms of the parameter t.
x= 1 −t
y= 1 + 2t
z=t
Step 6: The solution to the system of equations is:
x= 1 −t, y = 1 + 2t, z =t
29
Question 29
Question
Solve the following system of linear equations:
(2x−y=−1
4x+ 3y= 9
Solution
Step 1: Start by multiplying the first equation by 3 to eliminate y:
(6x−3y=−3
4x+ 3y= 9
Step 2: Add the two equations together to eliminate y:
10x= 6
Step 3: Solve for x:
x=6
10 =3
5
Step 4: Substitute x=3
5back into the first equation to solve for y:
23
5−y=−1
6
5−y=−1
y=6
5+ 1
y=6
5+5
5
y=11
5
Step 5: The solution to the system of equations is x=3
5and y=11
5.
Question 30
Question
Solve the following system of linear equations:
2x+ 3y−z= 5
x−2y+ 3z=−5
3x+y+z= 4
30
Solution
Step 1: Write the system of equations in augmented matrix form:
2 3 −1|5
1−2 3 | −5
3 1 1 |4
Step 2: Perform row operations to simplify the matrix.
2 3 −1|5
1−2 3 | −5
3 1 1 |4
→
1−2 3 | −5
2 3 −1|5
3 1 1 |4
Step 3: Use row operations to put the matrix in row-echelon form.
1−2 3 | −5
2 3 −1|5
3 1 1 |4
→
1−2 3 | −5
0 7 −7|15
0 7 −8|19
Step 4: Continue row operations to achieve reduced row-echelon form.
1−2 3 | −5
0 7 −7|15
0 7 −8|19
→
1−2 3 | −5
0 1 −1|15/7
0 0 −1|8/7
Step 5: Write the system of equations back in scalar form:
x−2y+ 3z=−5
y−z=15
7
−z=8
7
Step 6: Solve the system of equations to find x,y, and z. From the third
equation, we have z=−8
7. Substitute zinto the second equation: y−(−8
7) = 15
7,
so y= 1. Finally, substitute yand zinto the first equation: x−2(1) + 3(−8
7) =
−5, giving x=−11
7.
Therefore, the solution to the system of linear equations is:
x=−11
7, y = 1, z =−8
7
Question 31
Question
Solve the following system of linear equations using the method of substitution:
(3x−2y= 7
5x+ 4y= 11
31
Solution
Step 1: Solve the first equation for xin terms of y.
From the first equation, we have 3x−2y= 7, so 3x= 2y+ 7. Dividing by
3, we get x=2y+7
3.
Step 2: Substitute the expression for xinto the second equation.
Substitute x=2y+7
3into the second equation 5x+ 4y= 11:
52y+ 7
3+ 4y= 11
Step 3: Solve the resulting equation for y.
Multiplying both sides by 3 to get rid of the fraction:
5(2y+ 7) + 12y= 33
10y+ 35 + 12y= 33
22y+ 35 = 33
22y= 33 −35
22y=−2
y=−2
22
y=−1
11
Step 4: Substitute the value of yback into the equation we found for x.
Using x=2y+7
3, we get:
x=2(−1
11 )+7
3
x=−2/11 + 77/11
3
x=75
11 ·1
3
x=25
11
Step 5: Check the solutions by substituting them into the original equations.
Substitute x=25
11 and y=−1
11 into the original equations to verify the
solutions. Let’s check:
325
11−2−1
11= 7
525
11+ 4 −1
11= 11
Hence, the solutions are x=25
11 and y=−1
11 .
32
Question 32
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−3y+ 2z=−6
2x+y+z= 4
Solution
Step 1: Let’s rewrite the system of equations in matrix form:
3 2 −1
1−3 2
2 1 1
x
y
z
=
4
−6
4
Step 2: Now, we will perform row operations to row reduce the augmented
matrix:
3 2 −1 4
1−3 2 −6
2 1 1 4
Step 3: Subtract 2R1from R3to get the new R3:
3 2 −1 4
1−3 2 −6
0−3 3 −4
Step 4: Add R2to 3R3to get the new R3:
3 2 −1 4
1−3 2 −6
0 0 5 2
Step 5: Divide R3by 5 to simplify:
3 2 −1 4
1−3 2 −6
0 0 1 2
5
Step 6: Eliminate zfrom R1and R2:
32022
5
1−3 0 −16
5
001 2
5
Step 7: Eliminate yfrom R1:
1 0 0 4
010−2
001 2
5
Step 8: The solution to the system is x= 4, y=−2, z=2
5.
33
Question 33
Question
Solve the following system of linear equations:
2x−y+ 3z= 4
x+ 2y−z=−1
3x+ 4y+ 2z= 10
Solution
Step 1: Write down the augmented matrix for the system of equations.
2−1 3 |4
1 2 −1| −1
3 4 2 |10
Step 2: Perform row operations to get the matrix in row-echelon form.
1 2 −1| −1
0−5 5 |5
0−2 5 |13
Step 3: Continue row operations to get the matrix in reduced row-echelon
form.
1 0 −3| −7
0 1 −1| −1
0 0 0 |0
Step 4: Write down the system of equations represented by the reduced
row-echelon form. (x−3z=−7
y−z=−1
Step 5: Express the variables in terms of a parameter. Let z=t.
x=−7+3t
y=−1 + t
z=t
Step 6: Write the solution set in parametric vector form.
x
y
z
=
−7
−1
0
+t
3
1
1
34
Question 34
Question
Solve the following system of linear equations:
2x−3y+ 4z= 7
3x+ 2y−z=−5
x+ 4y−2z= 9
Solution
Step 1: Write the system of equations in matrix form:
2−3 4
3 2 −1
1 4 −2
x
y
z
=
7
−5
9
Step 2: Use row reduction to find the reduced row echelon form of the
augmented matrix:
2−3 4 7
3 2 −1−5
1 4 −2 9
1 4 −2 9
017
2
5
2−71
2
0−11 6 −14
1 4 −2 9
0 1 5
17 −71
34
0 0 101
34 −67
17
1 0 −3
17
95
34
0 1 5
17 −71
34
0 0 1 −134
101
Step 3: Write the system of equations represented by the row-echelon form:
x−3
17z=95
34
y+5
17z=−71
34
z=−134
101
Step 4: Solve the system of equations to find:
z=−134
101, y =−729
101, x =115
101
Therefore, the solution to the system of linear equations is:
x=115
101, y =−729
101, z =−134
101
35
Question 35
Question
Solve the following system of linear equations:
3x−2y+z= 7
2x+y−3z=−9
x−4y+ 2z= 1
Solution
Step 1: Write the system of equations in matrix form.
3−2 1
2 1 −3
1−4 2
x
y
z
=
7
−9
1
Step 2: Use Gaussian elimination to row-reduce the augmented matrix.
3−2 1 |7
2 1 −3| −9
1−4 2 |1
Step 3: Perform row operations to row-reduce the matrix:
1 0 1 |5
0 1 −5| −3
0 0 0 |0
Step 4: Solve the resulting system of equations:
(x+z= 5
y−5z=−3
Step 5: Solve for the variables: From the first equation: x= 5 −z
Substitute xinto the second equation: y−5z=−3
5−z−5z=−3
−6z=−8
z=4
3
x= 5 −4
3=11
3
y=−3+54
3=7
3
Therefore, the solution to the system of equations is x=11
3,y=7
3,z=4
3.
36
Question 3
Question
Solve the following system of linear equations:
2x−3y+ 5z= 4 (1)
4x+ 2y−z=−7 (2)
3x−4y+ 2z= 10 (3)
Solution
Step 1: Begin by rewriting 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.
2−3 5
4 2 −1
3−4 2
x
y
z
=
4
−7
10
Step 2: Use Gaussian elimination to row reduce the augmented matrix [A|b]
to row-echelon form:
2−3 5 |4
4 2 −1| −7
3−4 2 |10
Divide row 1 by 2:
1−3
2
5
2|2
4 2 −1| −7
3−4 2 |10
Subtract 4 times row 1 from row 2 and 3 times row 1 from row 3:
1−3
2
5
2|2
0 8 −11 | −15
0−1
2−1
2|4
Step 3: Continue row operations to further simplify the matrix to reduced
row-echelon form: Divide row 2 by 8 and row 3 by -0.5:
1−3
2
5
2|2
0 1 −11
8| −15
8
0 1 1 | −8
Subtract row 2 from row 3:
1−3
2
5
2|2
0 1 −11
8| −15
8
0 0 19
8| −97
8
3
Step 4: Solve for zusing the last row:
19
8z=−97
8=⇒z=−5
Step 5: Substitute z=−5 into the second row to solve for y:
y−11
8(−5) = −15
8=⇒y+55
8=−15
8=⇒y=−5
Step 6: Substitute z=−5 and y=−5 into the first row to solve for x:
x−3
2(−5) + 5
2(−5) = 2 =⇒x+15
2−25
2= 2 =⇒x−5 = 2 =⇒x= 7
Therefore, the solution to the given system of equations is x= 7, y=−5,
and z=−5.
Question 4
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
4x−y+ 3z=−4
x+ 2y−2z= 2
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
4−1 3
1 2 −2
, X =
x
y
z
, B =
1
−4
2
Step 2: Find the inverse of matrix Aby calculating A−1.
det(A) =
2 3 −1
4−1 3
1 2 −2
= 18
A−1=1
det(A)
A11 A21 A31
A12 A22 A32
A13 A23 A33
A−1=1
18
−8 9 1
−14 6 −2
−5 4 −2
4
Step 3: Solve for Xby multiplying X=A−1B.
X=
−8 9 1
−14 6 −2
−5 4 −2
1
−4
2
=
−17
−18
3
Therefore, the solution to the system of equations is x=−17, y=−18, and
z= 3.
Question 5
Question
Solve the following system of linear equations:
2x−3y+z= 4
x+ 2y−z=−1
3x−y+ 2z= 7
Solution
Step 1: Write the system of equations in matrix form as Ax=bwhere
A=
2−3 1
1 2 −1
3−1 2
,x=
x
y
z
,and b=
4
−1
7
Step 2: We want to find xsuch that Ax=b. Let’s find the inverse of
matrix Afirst.
Find det(A) = det
2−3 1
1 2 −1
3−1 2
= 29
Step 3: Find the adjugate matrix A∗of matrix A.
A∗=
3 7 5
−163
−5−8 4
T
=
3−1−5
7 6 −8
5 3 4
Step 4: Calculate the inverse of matrix Ausing the formula: A−1=
1
det(A)A∗.
A−1=1
29
3−1−5
7 6 −8
5 3 4
=
3
29 −1
29 −5
29
7
29
6
29 −8
29
5
29
3
29
4
29
Step 5: Multiply both sides of the equation Ax=bby A−1.
A−1(Ax) = A−1b
5
x=
3
29 −1
29 −5
29
7
29
6
29 −8
29
5
29
3
29
4
29
4
−1
7
Step 6: Perform matrix multiplication to find the values of x,y, and z.
This will give the solution to the system of equations.
Question 6
Question
Solve the following system of linear equations:
3x+ 2y+z= 7
x−y+ 2z= 3
2x+ 3y−z= 8
Solution
Step 1: Write the system of equations in matrix form as AX =B, where A
is the matrix of coefficients, Xis the column matrix of variables, and Bis the
column matrix of constants.
3 2 1
1−1 2
2 3 −1
x
y
z
=
7
3
8
Step 2: Find the inverse of matrix Ato solve for X.
A−1=1
det(A)adj(A)
First, calculate the determinant of matrix A:
det(A) = 3(−1)(−1) + 2(2)(2) + 1(1)(3) −1(−1)(2) −2(1)(3) −3(2)(−1) = 20
Step 3: Calculate the adjugate of matrix A.
adj(A) =
(−1)1+1M11 (−1)1+2M12 (−1)1+3M13
(−1)2+1M21 (−1)2+2M22 (−1)2+3M23
(−1)3+1M31 (−1)3+2M32 (−1)3+3M33
where Mij are the determinants of the (i, j) minors of matrix A.
Step 4: Calculate the inverse of matrix Ausing the formula A−1=1
det(A)adj(A).
A−1=1
20
6 1 5
5 3 1
−4−2 2
6
Step 5: Multiply both sides of the equation AX =Bby A−1to solve for X.
X=A−1B=1
20
6 1 5
5 3 1
−4−2 2
7
3
8
Step 6: Compute the product A−1Bto find the values of x,y, and z.
Question 7
Question
Find the solution to the system of linear equations:
3x−2y+z= 5
2x+y−3z=−1
x−4y+ 2z= 7
Solution
Step 1: Write the system of linear equations in matrix form as AX =B, where
Ais the coefficient matrix, Xis the variable matrix, and Bis the constant
matrix.
A=
3−2 1
2 1 −3
1−4 2
, X =
x
y
z
, B =
5
−1
7
Step 2: Find the inverse of matrix Aby calculating det(A) and A−1.
det(A) = 3(1(2) −(−3)(−4)) −(−2)(2(2) −(−3)(1)) + 1(2(−4) −1(−2)) = 47
A−1=1
det(A)
1(2) −(−3)(−4) −2(2) −(−3)(1) −2(−4) −1(−2)
−2(2) −(−3)(1) 3(2) −1(1) 1(−2) −3(−4)
1(−4) −2(−2) 2(−4) −1(1) 2(1) −2(2)
A−1=1
47
14 −7 6
−1 5 10
0−9−2
Step 3: Calculate the solution matrix X=A−1B.
X=1
47
14 −7 6
−1 5 10
0−9−2
5
−1
7
=1
47
(14 ·5) + (−7· −1) + (6 ·7)
(−1·5) + (5 · −1) + (10 ·7)
(0 ·5) + (−9· −1) + (−2·7)
7
X=1
47
70 + 7 + 42
−5−5 + 70
0+9−14
=1
47
119
60
23
Therefore, the solution to the system of linear equations is x=119
47 ,y=60
47 ,
and z=23
47 .
Question 8
Question
Solve the following system of linear equations:
2x−y+z= 5
x+ 3y−2z=−2
3x+y+ 4z= 10
Solution
Step 1: Rewrite the system of equations in matrix form:
2−1 1
1 3 −2
3 1 4
x
y
z
=
5
−2
10
Step 2: Use elementary row operations to row reduce the augmented matrix.
2−1 1 |5
1 3 −2| −2
3 1 4 |10
Step 3: Perform row operations to row reduce the matrix further.
1 3 −2| −2
0−5 5 |4
0−8 10 |16
Step 4: Continue row operations to simplify the matrix.
1 3 −2| −2
0 1 −1| −4/5
0 0 3 |12/5
Step 5: Simplify the equations back to variable form. From the third row,
3z=12
5⇒z=4
5. Substitute z=4
5back into the second row to solve for y:
y−4
5=−4
5⇒y= 0. Finally, substitute y= 0 and z=4
5into the first row to
solve for x: 2x−0 + 4
5= 5 ⇒x=23
10 .
Step 6: Therefore, the solution to the system of equations is x=23
10 , y =
0, z =4
5.
8
Question 9
Question
Solve the following system of linear equations:
2x−3y+z= 6
x+ 2y−2z=−8
3x−y+ 4z= 9
Solution
Step 1: Write the system of equations in matrix form:
2−3 1
1 2 −2
3−1 4
x
y
z
=
6
−8
9
Step 2: Row-reduce the augmented matrix:
2−3 1 6
1 2 −2−8
3−1 4 9
→
1 2 −2−8
0−1 5 22
0−7 10 33
→
1 2 −2−8
0 1 −5−22
0 0 45 121
Step 3: Perform back substitution to solve for z:
45z= 121 ⇒z=121
45 =121
45
Step 4: Substitute z=121
45 back into the second row to solve for y:
y−5121
45 =−22 ⇒y=−22 + 5 121
45 =77
9
Step 5: Substitute z=121
45 and y=77
9back into the first row to solve for
x:
x+ 2 77
9−2121
45 =−8⇒x=−8−154
9+242
45 =−111
5
Step 6: Therefore, the solution to the system of equations is x=−111
5,
y=77
9, and z=121
45 .
Question 10
Question
Find the values of xand ythat satisfy the following system of linear equations:
(3x−2y= 7
2x+ 4y= 1
9
Solution
To solve this system, we can use the method of substitution or the method of
elimination. Here, we will use the method of elimination.
Step 1: Multiply the first equation by 2 and the second equation by 3 to
create opposite coefficients for yin each equation.
(6x−4y= 14
6x+ 12y= 3
Step 2: Subtract the two equations to eliminate y.
−16y= 11
Step 3: Solve for y.
y=−11
16
Step 4: Substitute yback into one of the original equations to solve for x.
3x−2−11
16= 7
Step 5: Simplify and solve for x.
3x+22
16 = 7
3x= 7 −22
16
3x=82
16
x=82
48 =41
24
Step 6: Therefore, the solution to the system of equations is x=41
24 and
y=−11
16 .
Question 11
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
5x−y+ 3z= 2
2x+ 3y+z= 1
10
Solution
Step 1: We can write the system of equations in matrix form as Ax=bwhere
Ais the coefficient matrix, xis the column vector of variables, and bis the
column vector of constants.
3 2 −1
5−1 3
2 3 1
x
y
z
=
4
2
1
Step 2: To solve for x, we find the inverse of matrix A:
A−1=1
det(A)adj(A)
where det(A) is the determinant of Aand adj(A) is the adjugate of A.
Step 3: First, calculate the determinant of matrix A:
det(A) =
3 2 −1
5−1 3
2 3 1
=−59
Step 4: Next, find the matrix of cofactors, which is the transpose of the
matrix of minors:
cof(A) =
−8−11 13
1 5 −8
11 13 −9
Step 5: Calculate the adjugate by taking the transpose of the matrix of
cofactors:
adj(A) =
−8 1 11
−11 5 13
13 −8−9
Step 6: Now, find the inverse of matrix A:
A−1=1
−59
−8 1 11
−11 5 13
13 −8−9
=
8
59 −1
59 −11
59
11
59 −5
59 −13
59
−13
59
8
59
9
59
Step 7: Finally, solve for xby multiplying the inverse of matrix Awith b:
x
y
z
=
8
59 −1
59 −11
59
11
59 −5
59 −13
59
−13
59
8
59
9
59
4
2
1
After performing the matrix multiplication, we find the solution to be x=
−1, y = 2, z = 3.
11
Question 12
Question
Solve the following system of linear equations:
2x−3y+z= 7
4x+ 2y−3z=−2
x+ 5y−2z= 9
Solution
Step 1: Write the system of equations in augmented matrix form:
2−3 1 |7
4 2 −3| −2
1 5 −2|9
Step 2: Perform row operations to get leading 1s in the first column:
1−3
2
1
2|7
2
0 7 −4| −9
013
2−5
2|5
2
Step 3: Use row operations to get 0s below the leading 1 in the first column:
1−3
2
1
2|7
2
0 7 −4| −9
0 0 1 |3
Step 4: Continue row operations to get the reduced row-echelon form:
100|2
010| −1
001|3
Step 5: The solution to the system of equations is x= 2, y =−1, z = 3.
Question 13
Question
Solve the following system of linear equations using the method of elimination:
3x+ 2y−z= 7
2x−y+ 3z=−1
x+ 4y+ 2z= 18
12
Solution
Step 1: Multiply the first equation by 2 and the second equation by 3 to elimi-
nate xin the first two equations.
6x+ 4y−2z= 14
6x−3y+ 9z=−3
x+ 4y+ 2z= 18
Step 2: Subtract the second equation from the first equation to eliminate x.
6x+ 4y−2z= 14
−7y−11z=−17
x+ 4y+ 2z= 18
Step 3: Multiply the second equation by 4 and add it to the third equation
to eliminate y.
6x+ 4y−2z= 14
−7y−11z=−17
17z= 26
Step 4: Solve the third equation for z.
z=26
17 =2
17
Step 5: Substitute z=2
17 back into the second equation to solve for y.
−7y−11 2
17=−17
−7y−22
17 =−17
−7y=−17 + 22
17
−7y=−243
17
y=243
119
Step 6: Substitute y=243
119 and z=2
17 back into the first equation to solve
for x.
6x+ 4 243
119−22
17= 14
6x+972
119 −4
17 = 14
13
6x= 14 −972
119 +4
17
6x=260
119
x=130
119
Therefore, the solution to the system of linear equations is x=130
119 ,y=243
119 ,
and z=2
17 .
Question 14
Question
Solve the following system of linear equations:
2x+ 3y+z= 15
4x−y+ 2z=−1
x−2y+ 3z= 10
Solution
Step 1: Begin by writing the system of equations in matrix form:
231
4−1 2
1−2 3
x
y
z
=
15
−1
10
Step 2: To solve for x,y, and z, we will perform row operations to put the
system into row-echelon form.
Step 3: Replace R2with R2−2R1and R3with R3−1
2R1:
2 3 1
0−7 0
0−3.5 2.5
x
y
z
=
15
−31
2.5
Step 4: Next, replace R3with R3+ 0.5R2:
2 3 1
0−7 0
002.5
x
y
z
=
15
−31
−15.75
Step 5: Now, back substitute to find the values of x,y, and z:
2z=−15.75
−7y=−31
2x+ 3y+z= 15
Solving, we find z=−7.875, y= 4.429, and x= 3.718.
Therefore, the solution to the system of linear equations is x= 3.718, y=
4.429, and z=−7.875.
14
Question 15
Question
Solve the following system of linear equations:
2x−3y+z= 8
x+ 4y−2z=−2
3x−2y+ 5z= 10
Solution
Step 1: Start by writing the system of equations in matrix form:
2−3 1
1 4 −2
3−2 5
x
y
z
=
8
−2
10
Step 2: Use row operations to simplify the matrix:
2−3 1 8
1 4 −2−2
3−2 5 10
Step 3: Subtract 0.5×Row 1 from Row 2 and 1.5×Row 1 from Row 3:
2−3 1 8
0 5.5−2.5−6
0 2.5 3.5−2
Step 4: Divide Row 2 by 5.5 and subtract 2.5×Row 2 from Row 3:
2−3 1 8
0 1 −0.4545 −1.0909
0 0 4.3636 0.2727
Step 5: Divide Row 3 by 4.3636 and perform back substitution to find x,y,
and z:
2−3 1 8
0 1 −0.4545 −1.0909
0 0 1 0.0625
So, z= 0.0625.
Step 6: Substitute zback into Row 2 to find y:
y−0.4545(0.0625) = −1.0909
Solving for y, we get y=−1.
15
Step 7: Substitute yand zback into Row 1 to find x:
2x−3(−1) + 0.0625 = 8
Solving for x, we get x= 3.9375.
Therefore, the solution to the system of linear equations is:
x= 3.9375, y =−1, z = 0.0625
Question 16
Question
Consider the following system of linear equations:
2x−3y+z= 5
5x+ 2y−3z= 7
x−4y+ 2z= 1
Determine whether the system has a unique solution, no solution, or infinitely
many solutions. If a unique solution exists, solve the system.
Solution
Step 1: Write the system of equations in matrix form.
2−3 1
5 2 −3
1−4 2
x
y
z
=
5
7
1
Step 2: Calculate the determinant of the coefficient matrix.
det
2−3 1
5 2 −3
1−4 2
= 2(2(2)−(−3)(−4))−(−3)(5(2)−1(−4))+1(5(−4)−2(1))
= 2(4−12)−(−15 −(−4))+(20−2) = 2(−8)−(−11)+18 = −16 +11+18 = 13
Step 3: Since the determinant is non-zero (13 = 0), the system has a unique
solution.
Step 4: Use Cramer’s Rule to find the values of x, y, and z:
x=
det
5−3 1
7 2 −3
1−4 2
det
2−3 1
5 2 −3
1−4 2
=13
13 = 1
16
y=
det
2 5 1
5 7 −3
1 1 2
det
2−3 1
5 2 −3
1−4 2
=17
13
z=
det
2−3 5
527
1−4 1
det
2−3 1
5 2 −3
1−4 2
=23
13
Therefore, the system of equations has a unique solution: x= 1, y=17
13 ,
z=23
13 .
Question 17
Question
Solve the system of linear equations using the method of substitution:
(2x+ 3y= 7
4x−y= 5
Solution
Step 1: Solve the second equation for yin terms of x.
4x−y= 5
y= 4x−5
Step 2: Substitute the expression for yinto the first equation.
2x+ 3(4x−5) = 7
2x+ 12x−15 = 7
14x−15 = 7
Step 3: Solve for x.
14x−15 = 7
14x= 22
x=22
14
x=11
7
17
Step 4: Substitute the value of xback into the expression for yto find y.
y= 4 11
7−5
y=44
7−5
y=44
7−35
7
y=9
7
Therefore, the solution to the system of equations is:
(x=11
7
y=9
7
Question 18
Question
Solve the following system of linear equations:
3x+ 4y−z= 2
x−2y+ 3z=−4
2x+y= 1
Solution
Step 1: Write the system of equations as an augmented matrix:
3 4 −1 2
1−2 3 −4
2 1 0 1
Step 2: Perform row operations to get the matrix into row-echelon form.
1−2
301
3
010
3−1−10
3
05
311
3
Step 3: Continue row operations to get the matrix into reduced row-echelon
form.
1 0 2
5−1
5
0 1 −3
5−1
0 0 0 0
Step 4: Write the system of equations represented by the reduced matrix:
x+2
5z=−1
5
y−3
5z=−1
0 = 0
18
Step 5: Solve the system of equations. First, express yin terms of z:
y=3
5z−1
Step 6: Substitute yinto the equation for x:
x+2
5z=−1
5
x=−2
5z−1
5
Step 7: Therefore, the solution to the system of linear equations is:
x=−2
5z−1
5, y =3
5z−1, z =z
Question 19
Question
Solve the following system of linear equations:
2x−3y+z= 1
x+ 2y−z= 4
3x−y+ 2z= 5
Solution
Step 1: Write the system of equations as an augmented matrix.
2−3 1 1
1 2 −1 4
3−1 2 5
Step 2: Perform row operations to simplify the matrix.
2−3 1 1
1 2 −1 4
3−1 2 5
→
1 2 −1 4
2−3 1 1
3−1 2 5
Step 3: Perform row operations to further simplify the matrix.
1 2 −1 4
2−3 1 1
3−1 2 5
→
1 2 −1 4
0−7 3 −7
0−7 5 −7
Step 4: Continue performing row operations to row reduce the matrix.
1 2 −1 4
0−7 3 −7
0−7 5 −7
→
1 2 −1 4
0 1 −3/7 1
0−7 5 −7
→
1 2 −1 4
0 1 −3/7 1
0 0 26/7 0
19
Step 5: Convert the row reduced matrix back into a system of equations.
x+ 2y−z= 4
y−3
7z= 1
26
7z= 0
Step 6: Solve for the variables. From the third equation, we have z= 0.
Substituting this into the second equation, we get y−3
7(0) = 1, which simplifies
to y= 1. Finally, substituting z= 0 and y= 1 into the first equation, we get
x+ 2(1) −0 = 4, which simplifies to x= 2.
Therefore, the solution to the system of equations is x= 2, y= 1, and z= 0.
Question 20
Question
Solve the following system of linear equations:
2x−y+ 3z= 7
x+ 2y−z= 1
3x+y+ 2z= 11
Solution
Step 1: Rewrite the system of equations as an augmented matrix.
2−1 3 7
1 2 −1 1
3 1 2 11
Step 2: Perform row operations to obtain echelon form.
1 2 −1 1
0−5 4 6
0−5 5 8
Step 3: Continue row operations to reach reduced row-echelon form.
1 0 0 3
0 1 0 2
0 0 1 1
Step 4: Write the solution for the system of equations. The solution is x= 3,
y= 2, and z= 1.
20
Question 21
Question
Solve the following system of linear equations:
2x−3y+z= 7
x+ 2y−z=−1
3x−y= 6
Solution
Let’s solve the system of equations using the method of substitution.
Step 1: Let’s rewrite the system of equations:
2x−3y+z= 7 (1)
x+ 2y−z=−1 (2)
3x−y= 6 (3)
Step 2: From equation (3), express yin terms of x:
y= 3x−6
Step 3: Substitute y= 3x−6 into equation (2):
x+ 2(3x−6) −z=−1
x+ 6x−12 −z=−1
7x−z= 11 (4)
Step 4: Substitute y= 3x−6 into equation (1):
2x−3(3x−6) + z= 7
2x−9x+ 18 + z= 7
−7x+z=−11 (5)
Step 5: Now we have the system of equations:
7x−z= 11 (4)
−7x+z=−11 (5)
Step 6: Add equations (4) and (5) to eliminate z:
0=0
Since 0 = 0 is a true statement, the system of equations has infinitely many
solutions.
21
Question 22
Question
Solve the following system of linear equations:
2x−3y+z= 5
4x+y+ 2z= 4
3x+ 2y−z= 1
Solution
Step 1: Rewrite the system of equations in augmented matrix form.
2−315
4 1 2 4
3 2 −1 1
Step 2: Perform row operations to get zeros below the main diagonal.
2−3 1 5
0 7 0 −6
0 11 −4−14
Step 3: Perform row operations to get zeros above the main diagonal.
2−3 1 5
0 7 0 −6
0 0 −4 2
Step 4: Perform row operations to get a leading 1 in each row.
1−3
2
1
2
5
2
0 1 0 −6
7
0 0 1 −1
2
Step 5: Read off the solutions from the row-echelon form. The solution to
the system of equations is x=5
2,y=−6
7, and z=−1
2.
Question 23
Question
Find all values of ksuch that the system of linear equations is consistent:
2x−y+z= 3
4x+ky + 6z= 10
3x+ 2y+ 3z=k
22
Solution
Step 1: Rewrite the system of equations as an augmented matrix:
2−1 1 |3
4k6|10
323|k
Step 2: Perform row operations to simplify the matrix. First, replace the
second row with the result of subtracting twice the first row from it:
2−1 1 |3
0k+ 2 4 |4
323|k
Step 3: Next, replace the third row with the result of subtracting 3
2times
the first row from it:
2−1 1 |3
0k+ 2 4 |4
07
2
1
2|k−9
2
Step 4: Divide the third row by 7 to simplify the matrix further:
2−1 1 |3
0k+ 2 4 |4
0 1 1
2|k
7−9
14
Step 5: To ensure the system is consistent, the last row cannot have all zeros
on the left side while a non-zero value on the right. For this to happen, k+ 2
must be non-zero. Thus, we set up the equation k+ 2 = 0 and solve for k:
k+ 2 = 0
k=−2
Therefore, the system of linear equations is consistent for all values of k
except for k=−2.
Question 24
Question
Solve the system of linear equations:
2x−3y+ 4z= 7
x+y−z= 1
3x+ 2y−2z= 6
23
Solution
Step 1: Rewrite the system of equations in matrix form:
2−3 4
1 1 −1
3 2 −2
x
y
z
=
7
1
6
Step 2: Use row operations to row-reduce the augmented matrix:
2−347
1 1 −1 1
3 2 −2 6
R2=R2−1
2R1R3=R3−3
2R1
2−3 4 7
07
2−3−3
013
2−8−3
R3=R3−13
7R2
2−3 4 7
07
2−3−3
0 0 1 0
Step 3: Solve for z:7
2z−3 = −3
7
2z= 0
z= 0
Step 4: Substitute z= 0 back into the second row to solve for y:
7
2y−3 = −3
7
2y= 0
y= 0
Step 5: Substitute y= 0 and z= 0 back into the first row to solve for x:
2x−3(0) + 4(0) = 7
2x= 7
x=7
2
Therefore, the solution to the system of equations is x=7
2,y= 0, and
z= 0.
24
Question 25
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−y+z= 3
2x+ 3y+z= 0
Solution
Step 1: Write the system of linear equations in matrix form.
3 2 −1
1−1 1
2 3 1
x
y
z
=
4
3
0
Step 2: Use row operations to solve the system of equations by reducing
the augmented matrix to row-echelon form.
3 2 −1|4
1−1 1 |3
2 3 1 |0
Step 3: Subtract 2×Row 2 from Row 1:
1 4 −3| −2
1−1 1 |3
2 3 1 |0
Step 4: Subtract 2×Row 2 from Row 3:
1 4 −3| −2
1−1 1 |3
0 5 −1| −6
Step 5: Subtract 1/5×Row 3 from Row 2:
1 4 −3| −2
1 0 2 |4
0 1 −1/5|6/5
Step 6: Subtract 4×Row 3 from Row 1:
1 0 7/5| −22/5
1 0 2 |4
0 1 −1/5|6/5
25
Step 7: Simplify the augmented matrix and write the system of equations
back out: x+7
5z=−22
5
y−1
5z=6
5
Step 8: Solve for yin terms of zfrom the second equation:
y=1
5z+6
5
Step 9: Substitute yinto the first equation to solve for x:
x+7
5z=−22
5
x=−7
5z−22
5
Step 10: The general solution is:
x=−7
5z−22
5, y =1
5z+6
5, z is free
Question 26
Question
Consider the following system of linear equations:
2x+ 3y+z= 6
x−2y+ 2z= 5
3x+y−z= 1
Determine if the system has a unique solution, infinitely many solutions, or no
solution. If a unique solution exists, solve the system.
Solution
Step 1: Begin by writing the system of equations in matrix form:
2 3 1
1−2 2
3 1 −1
x
y
z
=
6
5
1
Step 2: Next, find the determinant of the coefficient matrix to determine the
type of solution the system has. The determinant Dis given by:
D=
2 3 1
1−2 2
3 1 −1
26
Step 3: Calculate the determinant using cofactor expansion along the first
row:
D= 2
−2 2
1−1
−3
1 2
3−1
+ 1
1−2
3 1
Step 4: Simplify the determinants and calculate D:
D= 2((−2)(−1) −(1)(2)) −3((1)(−1) −(3)(2)) + 1((1)(1) −(3)(−2))
D= 2(2 −2) −3(−1−6) + 1(1 + 6)
D= 0 + 21 + 7
D= 28
Step 5: Since D= 0, the system has a unique solution.
Step 6: Find the solution by calculating x,y, and zusing Cramer’s Rule:
x=Dx
D, y =Dy
D, z =Dz
D
where Dx,Dy, and Dzare the determinants obtained by replacing the x-column,
y-column, and z-column of the coefficient matrix with the constant vector.
Step 7: Calculate Dx,Dy, and Dz:
Dx=
6 3 1
5−2 2
1 1 −1
=−24
Dy=
2 6 1
1 5 2
3 1 −1
= 54
Dz=
236
1−2 5
311
=−32
Step 8: Calculate x,y, and z:
x=Dx
D=−24
28 =−6
7
y=Dy
D=54
28 =27
14
z=Dz
D=−32
28 =−8
7
Therefore, the system has a unique solution: x=−6
7,y=27
14 ,z=−8
7.
27
Question 27
Question
Solve the following system of linear equations:
2x−y+z= 3
4x+y−2z=−1
x+ 2y+z= 4
Solution
Step 1: Write the system of equations in matrix form:
2−1 1
4 1 −2
1 2 1
x
y
z
=
3
−1
4
Step 2: Use Gaussian elimination to row-reduce the augmented matrix:
2−1 1 |3
4 1 −2| −1
1 2 1 |4
R2−2R1
R3−1
2R1→
2−1 1 |3
0 3 −4| −7
05
2
1
2|5
2
R3−5
6R2→
2−1 1 |3
0 3 −4| −7
0 0 1 |3
Step 3: Back-substitute to find the values of x,y, and z: From the third
row, we have z= 3. Substitute z= 3 into the second row to find y:
3y−4(3) = −7
3y= 5
y=5
3
Substitute y=5
3and z= 3 into the first row to solve for x:
2x−5
3+ 3 = 3
2x=2
3
x=1
3
Therefore, the solution to the system of equations is x=1
3,y=5
3, and
z= 3.
28
Question 28
Question
Solve the following system of linear equations:
2x+ 3y−z= 1
4x−y+ 2z=−2
x+ 3y−3z= 0
Solution
Step 1: Write the system of equations in matrix form.
2 3 −1
4−1 2
1 3 −3
x
y
z
=
1
−2
0
Step 2: Use row reduction to find the reduced row echelon form of the
augmented matrix.
2 3 −1|1
4−1 2 | −2
1 3 −3|0
1 0 1 |1
0 1 −2|1
0 0 0 |0
Step 3: Write the system of equations corresponding to the reduced row
echelon form.
x+z= 1
y−2z= 1
0=0
Step 4: Solve the system of equations. From the last equation, we can see
that z is a free variable. Let z = t where t is a parameter.
Step 5: Express x and y in terms of the parameter t.
x= 1 −t
y= 1 + 2t
z=t
Step 6: The solution to the system of equations is:
x= 1 −t, y = 1 + 2t, z =t
29
Question 29
Question
Solve the following system of linear equations:
(2x−y=−1
4x+ 3y= 9
Solution
Step 1: Start by multiplying the first equation by 3 to eliminate y:
(6x−3y=−3
4x+ 3y= 9
Step 2: Add the two equations together to eliminate y:
10x= 6
Step 3: Solve for x:
x=6
10 =3
5
Step 4: Substitute x=3
5back into the first equation to solve for y:
23
5−y=−1
6
5−y=−1
y=6
5+ 1
y=6
5+5
5
y=11
5
Step 5: The solution to the system of equations is x=3
5and y=11
5.
Question 30
Question
Solve the following system of linear equations:
2x+ 3y−z= 5
x−2y+ 3z=−5
3x+y+z= 4
30
Solution
Step 1: Write the system of equations in augmented matrix form:
2 3 −1|5
1−2 3 | −5
3 1 1 |4
Step 2: Perform row operations to simplify the matrix.
2 3 −1|5
1−2 3 | −5
3 1 1 |4
→
1−2 3 | −5
2 3 −1|5
3 1 1 |4
Step 3: Use row operations to put the matrix in row-echelon form.
1−2 3 | −5
2 3 −1|5
3 1 1 |4
→
1−2 3 | −5
0 7 −7|15
0 7 −8|19
Step 4: Continue row operations to achieve reduced row-echelon form.
1−2 3 | −5
0 7 −7|15
0 7 −8|19
→
1−2 3 | −5
0 1 −1|15/7
0 0 −1|8/7
Step 5: Write the system of equations back in scalar form:
x−2y+ 3z=−5
y−z=15
7
−z=8
7
Step 6: Solve the system of equations to find x,y, and z. From the third
equation, we have z=−8
7. Substitute zinto the second equation: y−(−8
7) = 15
7,
so y= 1. Finally, substitute yand zinto the first equation: x−2(1) + 3(−8
7) =
−5, giving x=−11
7.
Therefore, the solution to the system of linear equations is:
x=−11
7, y = 1, z =−8
7
Question 31
Question
Solve the following system of linear equations using the method of substitution:
(3x−2y= 7
5x+ 4y= 11
31
Solution
Step 1: Solve the first equation for xin terms of y.
From the first equation, we have 3x−2y= 7, so 3x= 2y+ 7. Dividing by
3, we get x=2y+7
3.
Step 2: Substitute the expression for xinto the second equation.
Substitute x=2y+7
3into the second equation 5x+ 4y= 11:
52y+ 7
3+ 4y= 11
Step 3: Solve the resulting equation for y.
Multiplying both sides by 3 to get rid of the fraction:
5(2y+ 7) + 12y= 33
10y+ 35 + 12y= 33
22y+ 35 = 33
22y= 33 −35
22y=−2
y=−2
22
y=−1
11
Step 4: Substitute the value of yback into the equation we found for x.
Using x=2y+7
3, we get:
x=2(−1
11 )+7
3
x=−2/11 + 77/11
3
x=75
11 ·1
3
x=25
11
Step 5: Check the solutions by substituting them into the original equations.
Substitute x=25
11 and y=−1
11 into the original equations to verify the
solutions. Let’s check:
325
11−2−1
11= 7
525
11+ 4 −1
11= 11
Hence, the solutions are x=25
11 and y=−1
11 .
32
Question 32
Question
Solve the following system of linear equations:
3x+ 2y−z= 4
x−3y+ 2z=−6
2x+y+z= 4
Solution
Step 1: Let’s rewrite the system of equations in matrix form:
3 2 −1
1−3 2
2 1 1
x
y
z
=
4
−6
4
Step 2: Now, we will perform row operations to row reduce the augmented
matrix:
3 2 −1 4
1−3 2 −6
2 1 1 4
Step 3: Subtract 2R1from R3to get the new R3:
3 2 −1 4
1−3 2 −6
0−3 3 −4
Step 4: Add R2to 3R3to get the new R3:
3 2 −1 4
1−3 2 −6
0 0 5 2
Step 5: Divide R3by 5 to simplify:
3 2 −1 4
1−3 2 −6
0 0 1 2
5
Step 6: Eliminate zfrom R1and R2:
32022
5
1−3 0 −16
5
001 2
5
Step 7: Eliminate yfrom R1:
1 0 0 4
010−2
001 2
5
Step 8: The solution to the system is x= 4, y=−2, z=2
5.
33
Question 33
Question
Solve the following system of linear equations:
2x−y+ 3z= 4
x+ 2y−z=−1
3x+ 4y+ 2z= 10
Solution
Step 1: Write down the augmented matrix for the system of equations.
2−1 3 |4
1 2 −1| −1
3 4 2 |10
Step 2: Perform row operations to get the matrix in row-echelon form.
1 2 −1| −1
0−5 5 |5
0−2 5 |13
Step 3: Continue row operations to get the matrix in reduced row-echelon
form.
1 0 −3| −7
0 1 −1| −1
0 0 0 |0
Step 4: Write down the system of equations represented by the reduced
row-echelon form. (x−3z=−7
y−z=−1
Step 5: Express the variables in terms of a parameter. Let z=t.
x=−7+3t
y=−1 + t
z=t
Step 6: Write the solution set in parametric vector form.
x
y
z
=
−7
−1
0
+t
3
1
1
34
Question 34
Question
Solve the following system of linear equations:
2x−3y+ 4z= 7
3x+ 2y−z=−5
x+ 4y−2z= 9
Solution
Step 1: Write the system of equations in matrix form:
2−3 4
3 2 −1
1 4 −2
x
y
z
=
7
−5
9
Step 2: Use row reduction to find the reduced row echelon form of the
augmented matrix:
2−3 4 7
3 2 −1−5
1 4 −2 9
1 4 −2 9
017
2
5
2−71
2
0−11 6 −14
1 4 −2 9
0 1 5
17 −71
34
0 0 101
34 −67
17
1 0 −3
17
95
34
0 1 5
17 −71
34
0 0 1 −134
101
Step 3: Write the system of equations represented by the row-echelon form:
x−3
17z=95
34
y+5
17z=−71
34
z=−134
101
Step 4: Solve the system of equations to find:
z=−134
101, y =−729
101, x =115
101
Therefore, the solution to the system of linear equations is:
x=115
101, y =−729
101, z =−134
101
35
Question 35
Question
Solve the following system of linear equations:
3x−2y+z= 7
2x+y−3z=−9
x−4y+ 2z= 1
Solution
Step 1: Write the system of equations in matrix form.
3−2 1
2 1 −3
1−4 2
x
y
z
=
7
−9
1
Step 2: Use Gaussian elimination to row-reduce the augmented matrix.
3−2 1 |7
2 1 −3| −9
1−4 2 |1
Step 3: Perform row operations to row-reduce the matrix:
1 0 1 |5
0 1 −5| −3
0 0 0 |0
Step 4: Solve the resulting system of equations:
(x+z= 5
y−5z=−3
Step 5: Solve for the variables: From the first equation: x= 5 −z
Substitute xinto the second equation: y−5z=−3
5−z−5z=−3
−6z=−8
z=4
3
x= 5 −4
3=11
3
y=−3+54
3=7
3
Therefore, the solution to the system of equations is x=11
3,y=7
3,z=4
3.
36
