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 = $\left[ \begin{aligned} & \begin{matrix} x & 0 & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & x & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 0 & x \\\ \end{matrix} \\\ \end{aligned} \right]$ . 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 = $\left[ \begin{aligned} & \begin{matrix} x & 0 & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & x & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 0 & x \\\ \end{matrix} \\\ \end{aligned} \right]$ ………………………(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

& \Rightarrow 3x=12 \\\ & \Rightarrow x=\dfrac{12}{3} \\\ \end{aligned}$$ $$\Rightarrow x=4$$ …………………………..(4) Putting the value of x from equation (4) in equation (2), we get A = $$\left[ \begin{aligned} & \begin{matrix} 4 & 0 & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 4 & 0 \\\ \end{matrix} \\\ & \begin{matrix} 0 & 0 & 4 \\\ \end{matrix} \\\ \end{aligned} \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{aligned} & \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{aligned}$$ Therefore, the determinant value of matrix A is 64. Hence, the value of $$\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.