CBSE 2025 · Region 4 · Set 1 · Q32 · 5 marks
If A is a $\displaystyle 3 \times 3$ invertible matrix, show that for any scalar $\displaystyle \mathrm{k} \neq 0$, $\displaystyle (\mathrm{kA})^{-1}=\frac{1}{\mathrm{k}} \mathrm{A}^{-1}$. Hence calculate $\displaystyle (3 \mathrm{~A})^{-1}$, where \[\mathrm{A}=\left[\begin{array}{rrr} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \]
Marking-scheme solution
Consider \(\displaystyle (\mathrm{k} \mathrm{A})\left(\frac{1}{\mathrm{k}} \mathrm{A}^{-1}\right)=\mathrm{k} \cdot \frac{1}{\mathrm{k}}\left(\mathrm{A} \cdot \mathrm{A}^{-1}\right)=1 . I=I\)
\[\therefore(3 \mathrm{A})^{-1}=\frac{1}{3} \mathrm{A}^{-1}
\] Here, \(\displaystyle |\mathrm{A}|=4 \neq 0 \therefore \mathrm{A}^{-1}\) exists.
\[\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]
\]
\[\therefore(3 \mathrm{A})^{-1}=\frac{1}{3} \mathrm{A}^{-1}
\] Here, \(\displaystyle |\mathrm{A}|=4 \neq 0 \therefore \mathrm{A}^{-1}\) exists.
\[\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]
\]MatricesInvertible MatricesApplylong_answermedium
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