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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics4 Marks
Question
Find the inverse of the following matrices by the adjoint method : $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
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Answer

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

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