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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics5 Marks
Question
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right]$ find $A B$ and $(A B)^{-1}$. Verify that $(A B)^{-1}=B^{-1} A^{-1}$
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Answer

$\begin{aligned} A B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] & {\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] } =\left[\begin{array}{ll}2+9 & 0+3 \\ 1+6 & 0+2\end{array}\right]=\left[\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right]\end{aligned}$
$\therefore|A B|=\left|\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right|=22-21=1 \neq 0 \\ \therefore(A B)^{-1} $ exists. 
Now $,(A B)(A B)^{-1}=I $
$ \therefore\left[\begin{array}{rr}11 & 3 \\ 7 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right].$
By $2 R_1$, we get, $\left[\begin{array}{rr}22 & 6 \\ 7 & 2\end{array}\right](AB)^{-1}=\left[\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right]$
By $R_1-3 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 7 & 2\end{array}\right](A B)^{-1}=\left[\begin{array}{rr}2 & -3 \\ 0 & 1\end{array}\right]$
By $R_2-7 R_1$, we get, $\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right)(A B)^{-1}=\left[\begin{array}{rr}2 & -3 \\ -14 & 22\end{array}\right]$
By $\left(\frac{1}{2}\right) R_2$, we get,
$\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)(A B)^{-1}=\left(\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right)$
$\therefore(A B)^{-1}=\left(\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right)$
$|A|=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
Consider, $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_1 \leftrightarrow R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] A^{-1}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
By $R_2-2 R_1$, we get, $\left[\begin{array}{rr}1 & 2 \\ 0 & -1\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\ 1 & -2\end{array}\right]$
By $(-1) R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}0 & 1 \\ -1 & 2\end{array}\right]$
By $R_1-2 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right]$
$|B|=\left|\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right|=1-0=1 \neq 0$
$\therefore \mathrm{B}^{-1} $ exists. 
Consider, $\mathrm{BB}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}1 & 0 \\ 3 & 1\end{array}\right] \mathrm{B}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By $R_2-3 R_1$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] B^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right]$
$ \therefore B^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right] $
$ \therefore B^{-1} \cdot A^{-1}=\left[\begin{array}{rr}1 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{rr}2-0 & -3+0 \\ -6-1 & 9+2\end{array}\right] $
$ =\left[\begin{array}{rr}2 & -3 \\ -7 & 11\end{array}\right]$
From $(1)$ and $(2), (AB)^{-1} = B^{-1} ∙ A^{-1}.$

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