Exercise 4.6 class 12 maths ncert solutions | ex 4.6 class 12 maths ncert solutions | class 12 maths chapter 4 exercise 4.6 solutions | maths class 12 chapter 4 ncert solutions | determinants maths class 12 | class 12 maths determinants exercise 4.6
Looking for exercise 4.6 Class 12 Maths NCERT solutions? You’re in the right place! This section offers detailed, step-by-step answers to all the problems in Ex 4.6 Class 12 Maths NCERT Solutions, part of Chapter 4 – Determinants. In this exercise, you will learn to apply the concept of the inverse of a matrix using elementary row transformations, a vital technique in linear algebra. These Class 12 Maths Chapter 4 Exercise 4.6 Solutions follow the latest CBSE guidelines and are perfect for mastering determinants maths Class 12 topics. Whether you’re practicing for exams or need conceptual clarity, the Class 12 Maths Determinants Exercise 4.6 solutions are designed to support your learning journey effectively.
determinants maths class 12 || class 12 maths chapter 4 exercise 4.6 solutions || exercise 4.6 class 12 maths ncert solutions || ex 4.6 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || class 12 maths determinants exercise 4.6
Exercise 4.6
1. Examine the consistency of the system of equations.
\(\begin{array}{l}
x+2 y=2 \\
2 x+3 y=3
\end{array}\)
\(\begin{array}{l}
x+2 y=2 \\
2 x+3 y=3
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
x+2 y=2 \\
2 x+3 y=3
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}2 \\ 3\end{array}\right] \)
Now \( |A|=3(1)-2(2)=-1 \neq 0 \).
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
So, the system of equations will be consistent.
2. Examine the consistency of the system of equations.
\(\begin{array}{l}
2 x-y=5 \\
x+y=4
\end{array}\)
\(\begin{array}{l}
2 x-y=5 \\
x+y=4
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
2 x-y=5 \\
x+y=4
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}5 \\ 4\end{array}\right] \)
Now \( |\mathrm{A}|=2(1)-1(-1)=3 \neq 0 \).
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
So, the system of equations will be consistent.
3. Examine the consistency of the system of equations.
\(\begin{array}{l}
x+3 y=5 \\
2 x+6 y=8
\end{array}\)
\(\begin{array}{l}
x+3 y=5 \\
2 x+6 y=8
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
x+3 y=5 \\
2 x+6 y=8
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}5 \\ 8\end{array}\right] \)
Now \( |A|=1(6)-3(2)=0 \)
\( \therefore \mathrm{A} \) is a singular matrix and hence \( \mathrm{A}^{-1} \) doesn't exist.
So, the system of equations will be inconsistent.
determinants maths class 12 || class 12 maths chapter 4 exercise 4.6 solutions || exercise 4.6 class 12 maths ncert solutions || ex 4.6 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || class 12 maths determinants exercise 4.6
4. Examine the consistency of the system of equations.
\(x+y+z=1\)
\(\begin{array}{l}
2 x+3 y+2 z=2 \\
a x+a y+2 a z=4
\end{array}\)
\(x+y+z=1\)
\(\begin{array}{l}
2 x+3 y+2 z=2 \\
a x+a y+2 a z=4
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
x+y+z=1 \\
2 x+3 y+2 z=2 \\
a x+a y+2 a z=4
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right] \)
Now \( |A|=1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)=a \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
So the system of equations will be consistent.
5. Examine the consistency of the system of equations.
\(\begin{array}{l}
3 x-y-2 z=2 \\
2 y-z=-1 \\
3 x-5 y=3
\end{array}\)
\(\begin{array}{l}
3 x-y-2 z=2 \\
2 y-z=-1 \\
3 x-5 y=3
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
3 x-y-2 z=2 \\
0 x+2 y-z=-1
\end{array}\)
\(3 x-5 y+0 z=3\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right] \)
Now \( |\mathrm{A}|=3(-5)+3(5)=0 \)
\( \therefore \mathrm{A} \) is a singular matrix and hence \( \mathrm{A}^{-1} \) doesn't exist.
So the system of equations will be inconsistent.
6. Examine the consistency of the system of equations.
\(\begin{array}{l}
5 x-y+4 z=5 \\
2 x+3 y+5 z=2 \\
5 x-2 y+6 z=-1
\end{array}\)
\(\begin{array}{l}
5 x-y+4 z=5 \\
2 x+3 y+5 z=2 \\
5 x-2 y+6 z=-1
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
5 x-y+4 z=5 \\
2 x+3 y+5 z=2 \\
5 x-2 y+6 z=-1
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{c}5 \\ 2 \\ -1\end{array}\right] \)
Now \( |A|=5(18+10)+1(12-25)+4(-4-15)=51 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
So, the system of equations will be consistent.
7. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
5 x+2 y=4 \\
7 x+3 y=5
\end{array}\)
\(\begin{array}{l}
5 x+2 y=4 \\
7 x+3 y=5
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
5 x+2 y=4 \\
7 x+3 y=5
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}4 \\ 5\end{array}\right] \)
Now \( |A|=3(5)-2(7)=1 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}-1 \) exists.
Now \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A) \)
Adj \( A=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right] \)
So \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A)=\frac{1}{1}\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right] \)
And \( \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]\left[\begin{array}{l}4 \\ 5\end{array}\right]=\left[\begin{array}{c}2 \\ -3\end{array}\right] \)
So \( \left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ -3\end{array}\right] \)
Hence \( x=2 \) and \( y=-3 \).
8. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
2 x-y=-2 \\
3 x+4 y=3
\end{array}\)
\(\begin{array}{l}
2 x-y=-2 \\
3 x+4 y=3
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
2 x-y=-2 \\
3 x+4 y=3
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{c}-2 \\ 3\end{array}\right] \)
Now \( |A|=2(4)-3(-1)=11 \neq 0 \).
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}-1 \) exists.
Now \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A) \)
\( \operatorname{Adj} \mathrm{A}=\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right] \)
So \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A)=\frac{1}{11}=\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}\frac{4}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{2}{11}\end{array}\right] \)
And \( X=A^{-1} B=\left[\begin{array}{cc}\frac{4}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{2}{11}\end{array}\right]\left[\begin{array}{c}-2 \\ 3\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right] \)
So \( \left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right] \)
Hence \( x=-\frac{5}{11} \) and \( y=\frac{12}{11} \)
9. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
4 x-3 y=3 \\
3 x-5 y=7
\end{array}\)
\(\begin{array}{l}
4 x-3 y=3 \\
3 x-5 y=7
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
4 x-3 y=3 \\
3 x-5 y=7
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}3 \\ 7\end{array}\right] \)
Now \( |A|=4(-5)-3(-3)=-11 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}-1 \) exists.
Now \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A) \)
\( \operatorname{Adj} \mathrm{A}=\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right] \)
So \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A)=-\frac{1}{11}\left[\begin{array}{ll}-5 & 3 \\ -3 & 4\end{array}\right]=\left[\begin{array}{ll}\frac{5}{11} & \frac{-3}{11} \\ \frac{3}{11} & \frac{-4}{11}\end{array}\right] \)
And \( X=A^{-1} B=\left[\begin{array}{cc}\frac{5}{11} & \frac{-3}{11} \\ \frac{3}{11} & \frac{-4}{11}\end{array}\right]\left[\begin{array}{l}3 \\ 7\end{array}\right]=\left[\begin{array}{c}-\frac{6}{11} \\ -\frac{19}{11}\end{array}\right] \)
So \( \left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{r}-\frac{6}{11} \\ -\frac{19}{11}\end{array}\right] \)
Hence \( x=-\frac{6}{11} \) and \( y=\frac{-19}{11} \)
determinants maths class 12 || class 12 maths chapter 4 exercise 4.6 solutions || exercise 4.6 class 12 maths ncert solutions || ex 4.6 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || class 12 maths determinants exercise 4.6
10. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
5 x+2 y=3 \\
3 x+2 y=5
\end{array}\)
\(\begin{array}{l}
5 x+2 y=3 \\
3 x+2 y=5
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
5 x+2 y=3 \\
3 x+2 y=5
\end{array}\)
The given system of equations can be written in the form of \( A X=B \), where
\( \mathrm{A}=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}3 \\ 5\end{array}\right] \)
Now \( |A|=5(2)-2(3)=4 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
Now \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A) \)
\( \operatorname{Adj} A=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right] \)
So \( A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A)=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]=\left[\begin{array}{cc}\frac{2}{4} & \frac{-2}{4} \\ \frac{-3}{4} & \frac{5}{4}\end{array}\right] \)
And \( X=A^{-1} B=\left[\begin{array}{cc}\frac{1}{2} & \frac{-1}{2} \\ \frac{-3}{4} & \frac{5}{4}\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]=\left[\begin{array}{c}-1 \\ 4\end{array}\right] \)
So \( \left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ 4\end{array}\right] \)
Hence \( x=-1 \) and \( y=4 \)
11. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
2 x+y+z=1 \\
x-2 y-z=\frac{3}{2} \\
3 y-5 z=9
\end{array}\)
\(\begin{array}{l}
2 x+y+z=1 \\
x-2 y-z=\frac{3}{2} \\
3 y-5 z=9
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
2 x+y+z=1 \\
x-2 y-z=\frac{3}{2} \\
0 x+3 y-5 z=9
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}1 \\ \frac{3}{2} \\ 2 \\ 9\end{array}\right] \)
Now \( |\mathrm{A}|=2(10+3)-1(-5)+1(3)=34 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
Now \( A_{11}=13, A_{12}=5, A_{13}=3, A_{21}=8, A_{22}=-10, A_{23}=-6, A_{31}=1 \), \( \mathrm{A}_{32}=3, \mathrm{~A}_{33}=-5 \)
So Adj \( A=\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right] \)
\( \therefore A^{-1}=\frac{1}{|A|}(\operatorname{Adj} \mathrm{A})=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right] \)
And hence \( X=A^{-1} B \)
So
\( \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 5 & -6 & -5\end{array}\right]\left[\begin{array}{l}1 \\ \frac{3}{2} \\ 9\end{array}\right]=\frac{1}{34}\left[\begin{array}{c}34 \\ 17 \\ -51\end{array}\right]\)\(=\left[\begin{array}{c}\frac{34}{34} \\ \frac{17}{34} \\ -\frac{51}{34}\end{array}\right]=\left[\begin{array}{c}1 \\ \frac{1}{2} \\ -\frac{3}{2}\end{array}\right] \)
Hence \( x=1, y=\frac{1}{2} \) and \(z=-\frac{3}{2} \)
12. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
x-y+z=4 \\
2 x+y-3 z=0 \\
x+y+z=2
\end{array}\)
\(\begin{array}{l}
x-y+z=4 \\
2 x+y-3 z=0 \\
x+y+z=2
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
x-y+z=4 \\
2 x+y-3 z=0 \\
x+y+z=2
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right] \)
Now \( |A|=1(1+3)+1(5)+1(1)=10 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}-1 \) exists.
Now \( A_{11}=4, A_{12}=-5, A_{13}=1, A_{21}=2, A_{22}=0, A_{23}=-2, A_{31}=1, \) \( A_{32}=-2, \mathrm{~A}_{33}=3 \)
So Adj \( A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right] \)
\(\therefore A^{-1}=\frac{1}{|A|}(\operatorname{Adj} A)=\frac{1}{10}\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\)
And hence \( X=A^{-1} B \)
So
\( X=A^{-1} B=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]\)\(=\left[\begin{array}{c}\frac{20}{10} \\ \frac{-10}{10} \\ \frac{10}{10}\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right] \)
Hence \( x=2, y=-1 \) and \( z=1 \).
13. Solve system of linear equations, using matrix method.
\(\begin{array}{l}
2 x+3 y+3 z=5 \\
x-2 y+z=-4 \\
3 x-y-2 z=3
\end{array}\)
\(\begin{array}{l}
2 x+3 y+3 z=5 \\
x-2 y+z=-4 \\
3 x-y-2 z=3
\end{array}\)
Answer
The given system of equations is:
\(\begin{array}{l}
2 x+3 y+3 z=5 \\
x-2 y+z=-4 \\
3 x-y-2 z=3
\end{array}\)
The given system of equations can be written in the form of \( \mathrm{AX}=\mathrm{B} \), where
\( \mathrm{A}=\left[\begin{array}{ccc}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{array}\right] ; \mathrm{X}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \) and \( \mathrm{B}=\left[\begin{array}{c}5 \\ -4 \\ 3\end{array}\right] \)
Now \( |A|=2(4+1)-3(-5)+3(5)=40 \neq 0 \)
\( \therefore \mathrm{A} \) is a non-singular matrix and hence \( \mathrm{A}^{-1} \) exists.
Now \( A_{11}=5, A_{12}=5, A_{13}=5, A_{21}=3, A_{22}=-13, A_{23}=11, A_{31}=9, \) \(A_{32} =1, \mathrm{~A}_{33}=-7 \)
So Adj \( A=\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right] \)
\( \therefore \mathrm{A}^{-1}=\frac{1}{|A|}(\operatorname{Adj} \mathrm{A})=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right] \)
And hence \( X=A^{-1} B \)
So
\( X=A^{-1} B=\frac{1}{40}\left[\begin{array}{ccc}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{array}\right]\left[\begin{array}{c}5 \\ -4 \\ 3\end{array}\right]\)\(=\frac{1}{40}\left[\begin{array}{c}40 \\ 80 \\ -40\end{array}\right]=\left[\begin{array}{c}\frac{40}{40} \\ \frac{80}{40} \\ \frac{-40}{40}\end{array}\right]=\left[\begin{array}{c}1 \\ 2 \\ -1\end{array}\right] \)
Hence \( x=1, y=2 \) and \( z=-1 \)
