Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ is a $2 \times 2$ matrix and $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-\mathrm{x}+2$ is a polynomial, then what is $\mathrm{f}(\mathrm{A})$?
Options:
Solution:
2937 Upvotes
Verified Answer
The correct answer is:
$\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$
Given that, $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\left[\begin{array}{lr}1 & 2+6 \\ 0 & 9\end{array}\right]=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]$
Since, $f(x)=x^{2}-x+2$
Putting A in place of $x$ $f(A)=A^{2}-A+2 I$
$=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1-1+2 & 8-2+0 \\ 0-0+0 & 9-3+2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]=\left[\begin{array}{lr}1 & 2+6 \\ 0 & 9\end{array}\right]=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]$
Since, $f(x)=x^{2}-x+2$
Putting A in place of $x$ $f(A)=A^{2}-A+2 I$
$=\left[\begin{array}{ll}1 & 8 \\ 0 & 9\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1-1+2 & 8-2+0 \\ 0-0+0 & 9-3+2\end{array}\right]$
$=\left[\begin{array}{ll}2 & 6 \\ 0 & 8\end{array}\right]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.