Question

Adjoint of the matrix N=$\left[ \begin{matrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{matrix} \right]$ is [MP PET 1989]
E. N
F. 2N
G. - N
H. None of these

Answer 1

Hint: To answer the adjoint of a matrix question, we must first identify the cofactor of each matrix element. Create a new matrix using the cofactors, then expand the cofactors to produce the matrix. Then, transpose the matrix you determined in the previous step.

Formula Used: The following equation can be used to determine the cofactor for a specific element: $Aij = (-1)^{i+j} det M_{ij}$

Complete step by step solution:Given N=$\left[ \begin{matrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{matrix} \right]$
We will first evaluate each element's cofactor,
So, the cofactors of N are
${{c}_{11}}=-4,\,{{c}_{12}}=1,\,{{c}_{13}}=4$
${{c}_{21}}=-3,\,{{c}_{22}}=0,\,{{c}_{23}}=4$
${{c}_{31}}=-3,\,{{c}_{32}}=1,\,{{c}_{33}}=3$
Therefore, the transpose of the cofactor matrix is an Adjoint matrix.
N = $\left[ \begin{matrix} {{c}_{11}} & {{c}_{12}} & {{c}_{13}} \\ {{c}_{21}} & {{c}_{22}} & {{c}_{23}} \\ {{c}_{31}} & {{c}_{32}} & {{c}_{33}} \\ \end{matrix} \right]$=$\left[ \begin{matrix} 1 & 2 & -2 \\ 2 & 5 & -4 \\ 3 & 7 & -5 \\ \end{matrix} \right]$
Hence, the adjoint matrix formed is:
$adj\,N=\left[ \begin{matrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \\ \end{matrix} \right]$=N

Option ‘A’ is correct

Note: In a matrix—a cofactor is a number that is obtained by removing the row and column of a particular element. Generally, the cofactor is preceded by a positive (+) or negative (-) sign. Once the co-factor members of a matrix are transposed, the adjoint of the matrix is formed.