Browse Questions

# Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & 1& 3 \\ 4 &- 1 & 0 \\ -7 & 2 & 1 \\ \end{bmatrix}$

$\begin{array}{1 1} \frac{1}{-3}\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix} \\\frac{1}{3}\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix} \\-3\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix} \\ 3\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix} \end{array}$

Toolbox:
• (i)A matrix is said to be singular if $|A|$ =0.
• (ii)A matrix is said to be invertible only if $|A|\neq 0$.
• (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
• (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$[A_{ij}]n\times n$
• where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
Let A=$\begin{bmatrix}2 & 1& 3\\4 & -1& 0\\-7 & 2 & 1\end{bmatrix}$
Expanding along $R_1$ we get,
$|A|=2(-1\times 1-0\times 2)-1(4\times 1-(0\times -7)+3(4\times 2-(-7\times -1)$
$\;\;\;=2(-1)-1(4-0)+3(8-7)=-2-4+3=-3$
Since |A|$\neq 0$ inverse exists,
Now let us find the minors and cofactors of the elements.
$M_{11}=\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}=-1-0=-1$
$M_{12}=\begin{bmatrix}4 & 0\\-7 & 1\end{bmatrix}=4-0=4$
$M_{13}=\begin{bmatrix}4 & -1\\-7 & 2\end{bmatrix}=8-(7)=1$
$M_{21}=\begin{bmatrix}1 & 3\\2 & 1\end{bmatrix}=1-6=-5$
$M_{22}=\begin{bmatrix}2 & 3\\-7 & 1\end{bmatrix}=2+21=23$
$M_{23}=\begin{bmatrix}2 & 1\\-7 & 2\end{bmatrix}=4+7=11$
$M_{31}=\begin{bmatrix}1 & 3\\-1 & 0\end{bmatrix}=0+3=3$
$M_{32}=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}=0-12=-12$
$M_{33}=\begin{bmatrix}2 & 1\\4 & -1\end{bmatrix}=-2-4=-6$
$A_{11}=(-1)^{1+1}.(-1)=-1.$
$A_{12}=(-1)^{1+2}.(4)=-4.$
$A_{13}=(-1)^{1+3}.(1)=1.$
$A_{21}=(-1)^{2+1}.(-5)=5.$
$A_{22}=(-1)^{2+2}.(23)=23.$
$A_{23}=(-1)^{2+3}.11=-11.$
$A_{31}=(-1)^{3+1}.3=3.$
$A_{32}=(-1)^{3+2}.(-12)=12.$
$A_{33}=(-1)^{3+3}.-6=-6.$
Hence adj A=$\begin{bmatrix}A_{11} & A_{21}&A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} & A_{33}\end{bmatrix}$
$\;\;\;\;\qquad\quad=\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
|A|= -3
Therefore inverse =$\frac{1}{-3}\begin{bmatrix}-1 & 5 & 3\\-4 & 23 & 12\\1 & -11 & -6\end{bmatrix}$