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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics5 Marks
Question
Find $A^{-1}$ by adjoint method and by elementary transformations if $A=\left[\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$
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Answer

$|A|=\left|\begin{array}{lll}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{array}\right|$
$= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)$
$= 0 + 12 – 9 = 3 \neq 0$
$\therefore A^{-1}$ exists.
$A^{-1}$ by adjoint method :
We have to find the cofactor matrix
$= [A_{ij}]_{3\times 3}$​​​​​​​, where $A_{ij} = (-1)^{i+j}M_{ij}​​​​​​​$​​​​​​​
Now, $\mathrm{A}_{11}=(-1)^{1+1} \mathrm{M}_{11}=\left|\begin{array}{ll}1 & 2 \\ 2 & 4\end{array}\right|=4-4=0$
$A_{12}=(-1)^{1+2} M_{12}=-\left|\begin{array}{rrr}-1 & 2 \\ 1 & 4\end{array}\right|=-(-4-2)=6 $
$ A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{rr}-1 & 1 \\ 1 & 2\end{array}\right|=-2-1=-3 \\ A_{21}=(-1)^{2+1}$
$ M_{21}=-\left|\begin{array}{ll}2 & 3 \\ 2 & 4\end{array}\right|=-(8-6)=-2 \\ A_{22}=(-1)^{2+2} $
$M_{22}=\left|\begin{array}{ll}1 & 3 \\ 1 & 4\end{array}\right|=4-3=1$
$A_{23}=(-1)^{2+3} \mathrm{M}_{23}=-\left|\begin{array}{ll}1 & 2 \\ 1 & 2\end{array}\right|=-(2-2)=0 $
$ A_{31}=(-1)^{3+1} M_{31}=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1 $
$ A_{32}=(-1)^{3+2} \mathrm{M}_{32}=-\left|\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right|=-(2+3)=-5 $
$ A_{33}=(-1)^{3+3} \mathrm{M}_{33}=\left|\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right|=1+2=3$
$\therefore$ the cofactor matrix $=$
$\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}\end{array}\right]=\left[\begin{array}{rrr}0 & 6 & -3 \\-2 & 1 & 0 \\1 & -5 & 3\end{array}\right]$
$ \therefore \operatorname{adj} A  =\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{array}\right] $
$ \therefore A^{-1}  =\frac{1}{|A|}(\operatorname{adj} A) =\frac{1}{3}\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{array}\right]$
$\mathbf{A}^{-1}$ by elementary transformations :
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
By $R_2+R_1$ and $R_3-R_1$, we get,
$\therefore \left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
By $\left(\frac{1}{3}\right) R_2$, we get,
$\therefore\left[\begin{array}{rrr}1 & 2 & 3 \\0 & 1 & 5 / 3 \\0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}1 & 0 & 0 \\1 / 3 & 1 / 3 & 0 \\-1 & 0 & 1\end{array}\right]$
By $R_1-2 R_2$, we get,
$\therefore\left[\begin{array}{rrr}1 & 0 & -1 / 3 \\0 & 1 & 5 / 3 \\0 & 0 & 1\end{array}\right]  A^{-1}=\left[\begin{array}{rrr}1 / 3 & -2 / 3 & 0 \\1 / 3 & 1 / 3 & 0 \\-1 & 0 &1\end{array}\right]$
By $R_1+\frac{1}{3} R_3$ and $R_2-\frac{5}{3} R_3$, we get,
$\begin{array}{l}\therefore\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rrr}0 & -2 / 3 & 1 / 3 \\ 2 & 1 / 3 & -5 / 3 \\ -1 & 0 & 1\end{array}\right]\end{array} $
$ \therefore A^{-1}=\frac{1}{3}\left[\begin{array}{rrr}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3.\end{array}\right]$

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