Question

If A is a diagonal matrix of order \[3\times 3\] is a commutative with every square matrix of the order \[3\times 3\] under multiplication and tr(A)=12, then the value of \[\left| A \right|\] is

Answer 1

Hint: Assume the diagonal elements of the diagonal matrix A of the order \[3\times 3\] be x. We know that the diagonal matrix is such a matrix that has all elements equal to zero except the diagonal elements. Now, get the matrix A. We have the value of trace of matrix A and we know that trace of a matrix is the summation of its diagonal elements. Use this and get the value of x. Now, put the value of x in the matrix, A = \[\left[ \begin{align}
  & \begin{matrix}
   x & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & x & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & x \\
\end{matrix} \\
\end{align} \right]\] . Now, expand this matrix along the first row and get the determinant value of matrix A.

Complete step-by-step solution -
According to the question, it is given that if A is a diagonal matrix of order \[3\times 3\] is a commutative with every square matrix of the order \[3\times 3\] under multiplication and tr(A)= 12.
The trace of matrix A = 12 ……………………(1)
Let us assume the diagonal elements of the diagonal matrix A of the order \[3\times 3\] be x.
We know that the diagonal matrix is such a matrix that has all elements equal to zero except the diagonal elements.
Now, our diagonal matrix is,
A = \[\left[ \begin{align}
  & \begin{matrix}
   x & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & x & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & x \\
\end{matrix} \\
\end{align} \right]\] ………………………(2)
We know that trace of a diagonal matrix is the summation of the diagonal elements of the matrix.
From equation (2), we have the matrix A.
The summation of the diagonal elements of the matrix A = \[x+x+x=3x\] ………………………(3)
From equation (1), we have the trace of matrix A.
Now, from equation (2) and equation (3), we have
\[\begin{align}
  & \Rightarrow 3x=12 \\
 & \Rightarrow x=\dfrac{12}{3} \\
\end{align}\]
\[\Rightarrow x=4\] …………………………..(4)
Putting the value of x from equation (4) in equation (2), we get
A = \[\left[ \begin{align}
  & \begin{matrix}
   4 & 0 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 4 & 0 \\
\end{matrix} \\
 & \begin{matrix}
   0 & 0 & 4 \\
\end{matrix} \\
\end{align} \right]\] ……………………………..(5)
Now, we have to find the determinant value of matrix A.
From equation (5), we have matrix A.
Expanding the matrix A along the first row, we get
\[\begin{align}
  & \Rightarrow A=\left[ 4\left\{ 4\left( 4 \right)-0\left( 0 \right) \right\}-0\left\{ 0\left( 4 \right)-0\left( 0 \right) \right\}+0\left\{ 0\left( 0 \right)-0\left( 4 \right) \right\} \right] \\
 & \Rightarrow A=4\times 4\left( 4 \right) \\
 & \Rightarrow A=64 \\
\end{align}\]
Therefore, the determinant value of matrix A is 64.
Hence, the value of \[\left| A \right|\] is 64.

Note: In this question, one might take tr(A) as the transpose of matrix A. This is wrong because tr(A) means the trace of matrix A and trace of a matrix is the summation of its diagonal elements. To avoid this kind of silly mistake, we have to keep in mind that tr(A) means the trace of matrix A.