Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

Find the adjoint of the matrix $A =\begin{bmatrix}1&1&1\\2&1&-3\\-1&2&3\end{bmatrix}$.

Options:

$\begin{bmatrix}9&-1&-4\\-3&4&5\\5&-3&-1\end{bmatrix}$

$\begin{bmatrix}9&-1&-4\\3&-4&5\\5&-3&-1\end{bmatrix}$

$\begin{bmatrix}9&-1&-4\\-3&4&-5\\5&-3&-1\end{bmatrix}$

$\begin{bmatrix}-9&-1&-4\\-3&4&5\\-5&-3&-1\end{bmatrix}$

Correct Answer:

$\begin{bmatrix}9&-1&-4\\-3&4&5\\5&-3&-1\end{bmatrix}$

Explanation:

Let $C_{ij}$ be a cofactor of $a_{ij}$ in A. Then, the cofactors of elements of A are given by

$C_{11}=\begin{vmatrix}1&-3\\2&3\end{vmatrix}=9$

$C_{12}=-\begin{vmatrix}2&-3\\-1&3\end{vmatrix}=-3$

$C_{13}=\begin{vmatrix}2&1\\-1&2\end{vmatrix}=5$

$C_{21}=-\begin{vmatrix}1&1\\2&3\end{vmatrix}=-1$

$C_{22}=\begin{vmatrix}1&1\\-1&3\end{vmatrix}=4$

$C_{23}=-\begin{vmatrix}1&1\\-12&2\end{vmatrix}=-3$

$C_{31}=\begin{vmatrix}1&1\\-1&-3\end{vmatrix}=-4$

$C_{31}=-\begin{vmatrix}1&1\\2&-3\end{vmatrix}=5$

$C_{31}=\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1$

$∴adj\,A=\begin{bmatrix}9&-3&5\\-1&4&-3\\-4&5&-1\end{bmatrix}^T=\begin{bmatrix}9&-1&-4\\-3&4&5\\5&-3&-1\end{bmatrix}$