Find the inverses of the following matrices by the adjoint mathod…

Question

Find the inverses of the following matrices by the adjoint mathod : $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$

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Answer

Let $A =\left(\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right)$
Then $|A|=\left|\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right|$
$
\begin{aligned}
& =1(10-0)-2(0-0)+3(0-0) \\
& =10 \neq 0
\end{aligned}
$
$\therefore A ^{-1}$ exist. First we have to find the cofactor matrix $=\left[ A _{i j}\right]_{3 \times 3}$, where $A _{i j}=(-1)^{i+j} M _{i j}$
Now, $A_{11}=(-1)^{1+1} M_{11}=\left|\begin{array}{ll}2 & 4 \\ 0 & 5\end{array}\right|=10-0=10$
$
\begin{aligned}
& A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{ll}
0 & 4 \\
0 & 5
\end{array}\right|=-(0-0)=0 \\
& A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right|=0-0=0 \\
& A_{21}=(-1)^{2+1} M_{21}=-\left|\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right|=-(10-0)=-10
\end{aligned}
$
$
\begin{aligned}
& A_{22}=(-1)^{2+2} M_{22}=\left|\begin{array}{ll}
1 & 3 \\
0 & 5
\end{array}\right|=5-0=5 \\
& A_{23}=(-1)^{2+3} M_{23}=-\left|\begin{array}{ll}
1 & 2 \\
0 & 0
\end{array}\right|=-(0-0)=0 \\
& A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{ll}
2 & 3 \\
2 & 4
\end{array}\right|=8-6=2 \\
& A_{32}=(-1)^{3+2} M_{32}=-\left|\begin{array}{ll}
1 & 3 \\
0 & 4
\end{array}\right|=-(4-0)=-4 \\
& A_{33}=(-1)^{3+3} M_{33}=\left|\begin{array}{ll}
1 & 2 \\
0 & 2
\end{array}\right|=2-0=2
\end{aligned}
$
$\begin{aligned} & \therefore \text { the cofactor matrix } \\ & =\left(\begin{array}{lll} A _{11} & A _{12} & A _{13} \\ A _{21} & A _{22} & A _{23} \\ A _{31} & A _{32} & A _{33}\end{array}\right)=\left[\begin{array}{rrr}10 & 0 & 0 \\ -10 & 5 & 0 \\ 2 & -4 & 2\end{array}\right] \\ & \therefore \text { adj } A =\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right] \\ & \therefore A ^{-1}=\frac{1}{| A |} \text { (adj A) } \\ & \therefore A ^{-1}=\frac{1}{10}\left(\begin{array}{rrr}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]\end{aligned}$

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