Question

If $k$ is a scalar matrix and $I$is a unit matrix of order $3$then $adj(k\,I)=$.
A. ${{k}^{3}}I$
B. ${{k}^{2}}I$
C. $-{{k}^{3}}I$
D. $-{{k}^{2}}I$

Answer 1

Hint: Scalar matrix can be defined as the square matrix in which all the elements of the principal diagonal are some constant and all the other elements are zero.
Identity matrix can be defined as the square matrix in which all the elements of the principal diagonal are zero and all the other elements are zero. The identity matrix of order $3$is $I=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)$.

Complete step by step solution: We are given that $k$ is a scalar matrix and $I$is a unit matrix of order $3$and we have to find the value of $adj(k\,I)$.
We will take an identity matrix of order $3$ $I=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)$ and a scalar matrix $k$. Now we will multiply both the matrices.
$kI=k\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)$
$kI=\left( \begin{matrix}
   k & 0 & 0 \\
   0 & k & 0 \\
   0 & 0 & k \\
\end{matrix} \right)$
Now we will find the adjoint of the matrix.
$adj(kI)=\left( \begin{matrix}
   {{k}^{2}} & 0 & 0 \\
   0 & {{k}^{2}} & 0 \\
   0 & 0 & {{k}^{2}} \\
\end{matrix} \right)$
Taking ${{k}^{2}}$common from the matrix,
\[adj(kI)={{k}^{2}}\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
\[adj(kI)={{k}^{2}}I\]
The value of $adj(k\,I)$is \[adj(kI)={{k}^{2}}I\] where$k$ is a scalar matrix and $I$is a unit matrix of order $

Option ‘B’ is correct

Note: The relationship between the scalar matrix and the unit matrix is $Constant\times Identity\,\,matrix=Scalar\,matrix$.
All of the scalar matrices are symmetric in nature. The zero matrix is also a scalar matrix.
The adjoint of the matrix can be defined as the transpose of the cofactor of the matrix. Let us take a matrix of order $3$be$A=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$ . Then the transpose of this matrix will be${{A}^{T}}=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{21}} & {{a}_{31}} \\
   {{a}_{12}} & {{a}_{22}} & {{a}_{32}} \\
   {{a}_{13}} & {{a}_{23}} & {{a}_{33}} \\
\end{matrix} \right)$ where $\left( \begin{matrix}
   {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\
   {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\
   {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\
\end{matrix} \right)$ is the co-factor.
The relationship between the adjoint of the matrix and the identity matrix is $A\,(adj.A)=\,(adj.A).A=|A|I$.
For any of the scalar $k$, the adjoint will be $\,(adj\,kA)=\,{{k}^{n-1}}adj.A$.