CBSE 2023 · Region 1 · Set 1 · Q33 · 5 marks
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$ and $\displaystyle \mathrm{B}^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$, find $\displaystyle (\mathrm{AB})^{-1}$.Solve the following system of equations by matrix method : \[\begin{aligned} & x+2 y+3 z=6 \\ & 2 x-y+z=2 \\ & 3 x+2 y-2 z=3 \end{aligned} \]
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$ and $\displaystyle \mathrm{B}^{-1}=\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$, find $\displaystyle (\mathrm{AB})^{-1}$.
Solve the following system of equations by matrix method : \[\begin{aligned} & x+2 y+3 z=6 \\ & 2 x-y+z=2 \\ & 3 x+2 y-2 z=3 \end{aligned} \]
Marking-scheme solution
$$\begin{aligned}
& (\mathrm{a}) \mathrm{A}=\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right], \mathrm{B}^{-1}=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right] \\
& (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1} \\
& |\mathrm{~A}|=1(3)-2(-1)-2(2)=3+2-4=1 \neq 0
\end{aligned}
\therefore \mathrm{B}^{-1} \mathrm{~A}^{-1}=\left[\begin{array}{crr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]
$$
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