Examples (Revised) - Chapter 4 Determinants class 12 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Example 1

Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$.
Solution

We have $\left|\begin{array}{cc}2 & 4 \\ -1 & 2\end{array}\right|=2(2)-4(-1)=4+4=8$.
Example 2

Evaluate $\left|\begin{array}{cc}x & x+1 \\ x-1 & x\end{array}\right|$
Solution

We have
$
\left|\begin{array}{cc}
x & x+1 \\
x-1 & x
\end{array}\right|=x(x)-(x+1)(x-1)=x^2-\left(x^2-1\right)=x^2-x^2+1=1
$

Example 3

Evaluate the determinant $\Delta=\left|\begin{array}{rrr}1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0\end{array}\right|$.
Solution

Note that in the third column, two entries are zero. So expanding along third column $\left(\mathrm{C}_3\right)$, we get
$
\begin{aligned}
\Delta & =4\left|\begin{array}{rr}
-1 & 3 \\
4 & 1
\end{array}\right|-0\left|\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right|+0\left|\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right| \\
& =4(-1-12)-0+0=-52
\end{aligned}
$

Example 4

Evaluate 

$\Delta=\left|\begin{array}{ccc}
0 & \sin \alpha & -\cos \alpha \\
-\sin \alpha & 0 & \sin \beta \\
\cos \alpha & -\sin \beta & 0
\end{array}\right|$

Solution

Expanding along $\mathrm{R}_1$, we get
$
\begin{aligned}
\Delta & =0\left|\begin{array}{cc}
0 & \sin \beta \\
-\sin \beta & 0
\end{array}\right|-\sin \alpha\left|\begin{array}{cc}
-\sin \alpha & \sin \beta \\
\cos \alpha & 0
\end{array}\right|-\cos \alpha\left|\begin{array}{cc}
-\sin \alpha & 0 \\
\cos \alpha & -\sin \beta
\end{array}\right| \\
& =0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0) \\
& =\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0
\end{aligned}
$

Example 5

Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.
Solution

We have $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
i.e.
$
\begin{aligned}
3-x^2 & =3-8 \\
x^2 & =8 \\
x & = \pm 2 \sqrt{2}
\end{aligned}
$

Example 6

Find the area of the triangle whose vertices are $(3,8),(-4,2)$ and $(5,1)$. Solution The area of triangle is given by
$
\Delta=\frac{1}{2}\left|\begin{array}{rrr}
3 & 8 & 1 \\
-4 & 2 & 1 \\
5 & 1 & 1
\end{array}\right|
$

$
\begin{aligned}
& =\frac{1}{2}[3(2-1)-8(-4-5)+1(-4-10)] \\
& =\frac{1}{2}(3+72-14)=\frac{61}{2}
\end{aligned}
$

Example 7

Find the equation of the line joining $A(1,3)$ and $B(0,0)$ using determinants and find $k$ if $\mathrm{D}(k, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is 3 sq units.
Solution Let $\mathrm{P}(x, y)$ be any point on $\mathrm{AB}$. Then, area of triangle $\mathrm{ABP}$ is zero (Why?). So
$
\frac{1}{2}\left|\begin{array}{lll}
0 & 0 & 1 \\
1 & 3 & 1 \\
x & y & 1
\end{array}\right|=0
$

This gives
$
\frac{1}{2}(y-3 x)=0 \text { or } y=3 x,
$
which is the equation of required line $A B$.
Also, since the area of the triangle $\mathrm{ABD}$ is 3 sq. units, we have
$
\frac{1}{2}\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
k & 0 & 1
\end{array}\right|= \pm 3
$

This gives, $\frac{-3 k}{2}= \pm 3$, i.e., $k=\mp 2$.

Example 8

Find the minor of element 6 in the determinant $\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right|$
Solution

Since 6 lies in the second row and third column, its minor $\mathrm{M}_{23}$ is given by
$
\mathrm{M}_{23}=\left|\begin{array}{ll}
1 & 2 \\
7 & 8
\end{array}\right|=8-14=-6 \text { (obtained by deleting } \mathrm{R}_2 \text { and } \mathrm{C}_3 \text { in } \Delta \text { ). }
$

Example 9

Find minors and cofactors of all the elements of the determinant $\left|\begin{array}{rr}1 & -2 \\ 4 & 3\end{array}\right|$
Solution

Minor of the element $a_{i j}$ is $\mathrm{M}_{i j}$
Here $a_{11}=1$. So $\mathrm{M}_{11}=$ Minor of $a_{11}=3$
$\mathrm{M}_{12}=$ Minor of the element $a_{12}=4$
$\mathrm{M}_{21}=$ Minor of the element $a_{21}=-2$
$\mathrm{M}_{22}=$ Minor of the element $a_{22}=1$
Now, cofactor of $a_{i j}$ is $\mathrm{A}_{i j^{\prime}}$. So
$
\begin{array}{ll}
A_{11}=(-1)^{1+1} & M_{11}=(-1)^2(3)=3 \\
A_{12}=(-1)^{1+2} & M_{12}=(-1)^3(4)=-4 \\
A_{21}=(-1)^{2+1} & M_{21}=(-1)^3(-2)=2 \\
A_{22}=(-1)^{2+2} & M_{22}=(-1)^4(1)=1
\end{array}
$

Example 10.

Find minors and cofactors of the elements $a_{11}, a_{21}$ in the determinant
$
\Delta=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|
$

Solution

By definition of minors and cofactors, we have
Minor of $a_{11}=\mathrm{M}_{11}=\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|=a_{22} a_{33}-a_{23} a_{32}$
Cofactor of $a_{11}=\mathrm{A}_{11}=(-1)^{1+1} \quad \mathrm{M}_{11}=a_{22} a_{33}-a_{23} a_{32}$
Minor of $a_{21}=\mathrm{M}_{21}=\left|\begin{array}{ll}a_{12} & a_{13} \\ a_{32} & a_{33}\end{array}\right|=a_{12} a_{33}-a_{13} a_{32}$
Cofactor of $a_{21}=\mathrm{A}_{21}=(-1)^{2+1} \quad \mathrm{M}_{21}=(-1)\left(a_{12} a_{33}-a_{13} a_{32}\right)=-a_{12} a_{33}+a_{13} a_{32}$

Example 11

Find minors and cofactors of the elements of the determinant
$
\left|\begin{array}{ccc}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right| \text { and verify that } a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33}=0
$

Solution

We have $\mathrm{M}_{11}=\left|\begin{array}{cc}0 & 4 \\ 5 & -7\end{array}\right|=0-20=-20 ; \mathrm{A}_{11}=(-1)^{1+1}(-20)=-20$
$
\begin{array}{ll}
\mathrm{M}_{12}=\left|\begin{array}{cc}
6 & 4 \\
1 & -7
\end{array}\right|=-42-4=-46 ; & \mathrm{A}_{12}=(-1)^{1+2}(-46)=46 \\
\mathrm{M}_{13}=\left|\begin{array}{ll}
6 & 0 \\
1 & 5
\end{array}\right|=30-0=30 ; & \mathrm{A}_{13}=(-1)^{1+3}(30)=30
\end{array}
$

$
\begin{array}{ll}
\mathrm{M}_{21}=\left|\begin{array}{cc}
-3 & 5 \\
5 & -7
\end{array}\right|=21-25=-4 ; & \mathrm{A}_{21}=(-1)^{2+1}(-4)=4 \\
\mathrm{M}_{22}=\left|\begin{array}{cc}
2 & 5 \\
1 & -7
\end{array}\right|=-14-5=-19 ; & \mathrm{A}_{22}=(-1)^{2+2}(-19)=-19 \\
\mathrm{M}_{23}=\left|\begin{array}{cc}
2 & -3 \\
1 & 5
\end{array}\right|=10+3=13 ; & \mathrm{A}_{23}=(-1)^{2+3}(13)=-13 \\
\mathrm{M}_{31}=\left|\begin{array}{cc}
-3 & 5 \\
0 & 4
\end{array}\right|=-12-0=-12 ; & \mathrm{A}_{31}=(-1)^{3+1}(-12)=-12 \\
\mathrm{M}_{32}=\left|\begin{array}{ll}
2 & 5 \\
6 & 4
\end{array}\right|=8-30=-22 ; & \mathrm{A}_{32}=(-1)^{3+2}(-22)=22 \\
\mathrm{M}_{33}=\left|\begin{array}{cc}
2 & -3 \\
6 & 0
\end{array}\right|=0+18=18 ; & \mathrm{A}_{33}=(-1)^{3+3}(18)=18
\end{array}
$
and
Now
$
a_{11}=2, a_{12}=-3, a_{13}=5 ; \mathrm{A}_{31}=-12, \mathrm{~A}_{32}=22, \mathrm{~A}_{33}=18
$

So
$
\begin{aligned}
& a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33} \\
& =2(-12)+(-3)(22)+5(18)=-24-66+90=0
\end{aligned}
$

Example 12

Find $\operatorname{adj} \mathrm{A}$ for $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right]$
Solution

We have $\mathrm{A}_{11}=4, \mathrm{~A}_{12}=-1, \mathrm{~A}_{21}=-3, \mathrm{~A}_{22}=2$
Hence
$
\operatorname{adj} \mathrm{A}=\left[\begin{array}{ll}
\mathrm{A}_{11} & \mathrm{~A}_{21} \\
\mathrm{~A}_{12} & \mathrm{~A}_{22}
\end{array}\right]=\left[\begin{array}{cc}
4 & -3 \\
-1 & 2
\end{array}\right]
$

Example 13

If $\mathrm{A}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$, then verify that $\mathrm{A}$ adj $\mathrm{A}=|\mathrm{A}| \mathrm{I}$. Also find $\mathrm{A}^{-1}$.
Solution

We have $|A|=1(16-9)-3(4-3)+3(3-4)=1 \neq 0$
Now $\mathrm{A}_{11}=7, \mathrm{~A}_{12}=-1, \mathrm{~A}_{13}=-1, \mathrm{~A}_{21}=-3, \mathrm{~A}_{22}=1, \mathrm{~A}_{23}=0, \mathrm{~A}_{31}=-3, \mathrm{~A}_{32}=0$, $\mathrm{A}_{33}=1$

Therefore
$
\operatorname{adj} \mathrm{A}=\left[\begin{array}{rrr}
7 & -3 & -3 \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right]
$

Now
$
\begin{aligned}
\mathrm{A}(\operatorname{adj} \mathrm{A}) & =\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\left[\begin{array}{rrr}
7 & -3 & -3 \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rrr}
7-3-3 & -3+3+0 & -3+0+3 \\
7-4-3 & -3+4+0 & -3+0+3 \\
7-3-4 & -3+3+0 & -3+0+4
\end{array}\right] \\
& =\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=(1) \quad\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=|\mathrm{A}| \cdot \mathrm{I} \\
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \text { adj A } & =\frac{1}{1}\left[\begin{array}{rrr}
7 & -3 & -3 \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
7 & -3 & -3 \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right]
\end{aligned}
$

Example 14

If $A=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]$, then verify that $(A B)^{-1}=B^{-1} A^{-1}$.
Solution

We have $A B=\left[\begin{array}{cc}2 & 3 \\ 1 & -4\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]=\left[\begin{array}{cc}-1 & 5 \\ 5 & -14\end{array}\right]$
Since, $|A B|=-11 \neq 0,(A B)^{-1}$ exists and is given by
$
(\mathrm{AB})^{-1}=\frac{1}{|\mathrm{AB}|} \operatorname{adj}(\mathrm{AB})=-\frac{1}{11}\left[\begin{array}{cc}
-14 & -5 \\
-5 & -1
\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}
14 & 5 \\
5 & 1
\end{array}\right]
$

Further, $|A|=-11 \neq 0$ and $|B|=1 \neq 0$. Therefore, $A^{-1}$ and $B^{-1}$ both exist and are given by
$
\mathrm{A}^{-1}=-\frac{1}{11}\left[\begin{array}{cc}
-4 & -3 \\
-1 & 2
\end{array}\right], \mathrm{B}^{-1}=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]
$

Therefore
$
\mathrm{B}^{-1} \mathrm{~A}^{-1}=-\frac{1}{11}\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{cc}
-4 & -3 \\
-1 & 2
\end{array}\right]=-\frac{1}{11}\left[\begin{array}{cc}
-14 & -5 \\
-5 & -1
\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}
14 & 5 \\
5 & 1
\end{array}\right]
$

Hence $(A B)^{-1}=B^{-1} A^{-1}$

Example 15

Show that the matrix $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ satisfies the equation $A^2-4 A+I=O$, where $\mathrm{I}$ is $2 \times 2$ identity matrix and $\mathrm{O}$ is $2 \times 2$ zero matrix. Using this equation, find $\mathrm{A}^{-1}$.

Solution

We have $A^2=A . A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]$
Hence
$
\mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]-\left[\begin{array}{cc}
8 & 12 \\
4 & 8
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O}
$

Now
$
\mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\mathrm{O}
$

Therefore
$\mathrm{AA}-4 \mathrm{~A}=-\mathrm{I}$
or
A A $\left(A^{-1}\right)-4 \mathrm{AA}^{-1}=-\mathrm{IA}^{-1}$ (Post multiplying by $\mathrm{A}^{-1}$ because $|\mathrm{A}| \neq 0$ )

or
$
A\left(A^{-1}\right)-4 I=-A^{-1}
$
or
$
A I-4 I=-A^{-1}
$
or
$
A^{-1}=4 I-A=\left[\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right]-\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]
$

Hence
$
A^{-1}=\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]
$

Example 16

Solve the system of equations
$
\begin{aligned}
& 2 x+5 y=1 \\
& 3 x+2 y=7
\end{aligned}
$

Solution

The system of equations can be written in the form $\mathrm{AX}=\mathrm{B}$, where
$
\mathbf{A}=\left[\begin{array}{ll}
2 & 5 \\
3 & 2
\end{array}\right], \mathbf{X}=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } \mathbf{B}=\left[\begin{array}{l}
1 \\
7
\end{array}\right]
$

Now, $|A|=-11 \neq 0$, Hence, $A$ is nonsingular matrix and so has a unique solution.

Note that
$
\mathrm{A}^{-1}=-\frac{1}{11}\left[\begin{array}{cc}
2 & -5 \\
-3 & 2
\end{array}\right]
$

Therefore
$
\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=-\frac{1}{11}\left[\begin{array}{cc}
2 & -5 \\
-3 & 2
\end{array}\right]\left[\begin{array}{l}
1 \\
7
\end{array}\right]
$
i.e.
$
\left[\begin{array}{l}
x \\
y
\end{array}\right]=-\frac{1}{11}\left[\begin{array}{c}
-33 \\
11
\end{array}\right]=\left[\begin{array}{c}
3 \\
-1
\end{array}\right]
$

Hence
$x=3, y=-1$

Example 17

Solve the following system of equations by matrix method.
$
\begin{array}{r}
3 x-2 y+3 z=8 \\
2 x+y-z=1 \\
4 x-3 y+2 z=4
\end{array}
$

Solution

The system of equations can be written in the form $A X=B$, where
$
\mathrm{A}=\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]
$

We see that
$
|A|=3(2-3)+2(4+4)+3(-6-4)=-17 \neq 0
$

Hence, $\mathrm{A}$ is nonsingular and so its inverse exists. Now

$\begin{array}{lll}
A_{11}=-1, & A_{12}=-8, & A_{13}=-10 \\
A_{21}=-5, & A_{22}=-6, & A_{23}=1 \\
A_{31}=-1, & A_{32}=9, & A_{33}=7
\end{array}$

Therefore
$
\begin{aligned}
& A^{-1}=-\frac{1}{17}\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right] \\
& \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=-\frac{1}{17}\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right] \\
&
\end{aligned}
$
$
\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{l}
-17 \\
-34 \\
-51
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]
$

Hence
$x=1, y=2$ and $z=3$.

Example 18

The sum of three numbers is 6 . If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.
Solution

Let first, second and third numbers be denoted by $x, y$ and $z$, respectively. Then, according to given conditions, we have
$
\begin{aligned}
x+y+z & =6 \\
y+3 z & =11 \\
x+z & =2 y \text { or } x-2 y+z=0
\end{aligned}
$

This system can be written as $\mathrm{A} X=\mathrm{B}$, where
$
\mathrm{A}=\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & 2 & 1
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]
$

Here $|\mathrm{A}|=1(1+6)-(0-3)+(0-1)=9 \neq 0$. Now we find $a d j \mathrm{~A}$

$\begin{array}{lll}
A_{11}=1(1+6)=7, & A_{12}=-(0-3)=3, & A_{13}=-1 \\
A_{21}=-(1+2)=-3, & A_{22}=0, & A_{23}=-(-2-1)=3 \\
A_{31}=(3-1)=2, & A_{32}=-(3-0)=-3, & A_{33}=(1-0)=1
\end{array}$

$\text { Hence } \quad a d j \mathrm{~A}=\left[\begin{array}{ccc}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]$

Thus
$
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj}(\mathrm{A})=\frac{1}{9}\left[\begin{array}{ccc}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]
$

Since
$
\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}
$
$
\mathrm{X}=\frac{1}{9}\left[\begin{array}{ccc}
7 & -3 & 2 \\
3 & 0 & -3 \\
-1 & 3 & 1
\end{array}\right]\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]
$
or
$
\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{9}\left[\begin{array}{l}
42-33+0 \\
18+0+0 \\
-6+33+0
\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}
9 \\
18 \\
27
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]
$

Thus
$
x=1, y=2, z=3
$

Example 19

Use product $\left[\begin{array}{ccc}1 & 1 & 2 \\ 0 & 2 & 3 \\ 3 & 2 & 4\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 1 \\ 9 & 2 & 3 \\ 6 & 1 & 2\end{array}\right]$ to solve the system of equations
$
\begin{array}{r}
x-y+2 z=1 \\
2 y-3 z=1 \\
3 x-2 y+4 z=2
\end{array}
$

Solution

Consider the product $\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$
$
=\left[\begin{array}{ccc}
-2-9+12 & 0-2+2 & 1+3-4 \\
0+18-18 & 0+4-3 & 0-6+6 \\
-6-18+24 & 0-4+4 & 3+6-8
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$

Hence
$
\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]^{-1}=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]
$

Now, given system of equations can be written, in matrix form, as follows
$
\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]
$

or
$
\begin{aligned}
& x \\
& y=\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]^{-1}\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]=\left[\begin{array}{lll}
2 & 0 & 1 \\
9 & 2 & 3 \\
6 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
&=\left[\begin{array}{r}
-2+0+2 \\
9+2-6 \\
6+1-4
\end{array}\right]=\left[\begin{array}{l}
0 \\
5 \\
3
\end{array}\right]
\end{aligned}
$

Hence
$x=0, y=5$ and $z=3$