Question

If $A$ is a diagonal matrix of order $3 \times 3$is cumulative with every square matrix of order $3 \times 3$ under multiplication and $tr\left( A \right) = 12$, then the value of ${\left| A \right|^{1/2}}$ is …………

Answer 1

Hint: A diagonal matrix is one whose all non-diagonal elements are zero. For ex- $\left[ {\begin{array}{*{20}{c}}
  a&0&0 \\
  0&b&0 \\
  0&0&c
\end{array}} \right]$ is a diagonal matrix. Also, the trace of a matrix is the sum of the diagonal elements. For ex- the trace of the above matrix is \[a + b + c\].

Complete step-by-step answer:
 As we have given that the matrix $A$ is a diagonal matrix, it means that the matrix has a non-zero diagonal and the rest other elements are $0$.
Let $A = \left[ {\begin{array}{*{20}{c}}
  x&0&0 \\
  0&x&0 \\
  0&0&x
\end{array}} \right]$ ….. (1)
Here we have taken all the diagonal elements the same because it is given that the matrix $A$ is cumulative with every square matrix.
We know that the trace of a matrix is the sum of the diagonal elements.
Given, $tr\left( A \right) = 12$
$\therefore x + x + x = 12$
$ \Rightarrow 3x = 12$
$ \Rightarrow x = \dfrac{{12}}{3}$
$ \Rightarrow x = 4$
From (1);
$A = \left[ {\begin{array}{*{20}{c}}
  4&0&0 \\
  0&4&0 \\
  0&0&4
\end{array}} \right]$
Now solving this matrix as determinant and expand along ${R_1}$,
$\left| A \right| = 4\left( {4 \times 4 - 0} \right)$
$ \Rightarrow \left| A \right| = 64$
Hence, ${\left| A \right|^{1/2}} = \sqrt {64} = 8$
$\therefore {\left| A \right|^{1/2}} = 8$

Note: If a diagonal matrix of order $3 \times 3$is cumulative with every square matrix of order $3 \times 3$ under multiplication, then all the diagonal elements of the diagonal matrix are the same.