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

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.