Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right) A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, then the matrix $a$ is
MathematicsMatricesKCETKCET 2020
Options:
  • A $\left(\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right)$
  • B $\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
  • C $\left(\begin{array}{cc}-2 & 1 \\ 3 & -2\end{array}\right)$
  • D $\left(\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right)$
Solution:
1967 Upvotes Verified Answer
The correct answer is: $\left(\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right)$
We have,
$\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Let
$B=\left[\begin{array}{ll}
2 & 1 \\
3 & 2
\end{array}\right]$
$\therefore \quad B A=I$
$A=B^{-1} I$
$A=B^{-1}$
$\therefore \quad B^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
Hence, $A=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$

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.