Question

# If A is a diagonal matrix of order n such that If A is a diagonal matrix of order n such that ${{A}^{2}}=A$, then the number of possible values of matrix A is a) 0b) 1c) ${{2}^{n}}$ d) ${{2}^{n-1}}$

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Hint: In the diagonal matrix, all the elements of the matrix are zero except the diagonal running from the upper left to the lower left.
Let the elements of diagonal be ${{a}_{ii}}$ where i = 1,2,3…. n
Substitute the values in the diagonal and solve for ${{A}^{2}}=A$. Simplify the result to get a relation between the result and the diagonal elements and hence calculate the number of values for matrix A.

Consider a diagonal matrix A of order n:
$A=\left[ \begin{matrix} {{a}_{11}} & 0 & . & . \\ 0 & {{a}_{22}} & . & . \\ . & . & . & . \\ 0 & . & . & {{a}_{nn}} \\ \end{matrix} \right]$
where $\left( {{a}_{11}},{{a}_{22}},.....,{{a}_{nn}} \right)$ are elements of diagonal.
So,
${{A}^{2}}=\left[ \begin{matrix} {{a}_{11}} & 0 & . & . \\ 0 & {{a}_{22}} & . & . \\ . & . & . & . \\ 0 & . & . & {{a}_{nn}} \\ \end{matrix} \right]\left[ \begin{matrix} {{a}_{11}} & 0 & . & . \\ 0 & {{a}_{22}} & . & . \\ . & . & . & . \\ 0 & . & . & {{a}_{nn}} \\ \end{matrix} \right]$
${{A}^{2}}=\left[ \begin{matrix} {{\left( {{a}_{11}} \right)}^{2}} & 0 & . & . \\ 0 & {{\left( {{a}_{22}} \right)}^{2}} & . & . \\ . & . & . & . \\ 0 & . & . & {{\left( {{a}_{nn}} \right)}^{2}} \\ \end{matrix} \right]......(1)$
Hence, ${{A}^{2}}={{\left( \text{Diagonal element} \right)}^{2}}$
i.e. ${{A}^{2}}=\left[ {{\left( {{a}_{11}} \right)}^{2}},{{\left( {{a}_{22}} \right)}^{2}},......,{{\left( {{a}_{nn}} \right)}^{2}} \right]......(2)$

Since it is given in the question that ${{A}^{2}}=A$
Therefore, $\left[ {{\left( {{a}_{11}} \right)}^{2}},{{\left( {{a}_{22}} \right)}^{2}},......,{{\left( {{a}_{nn}} \right)}^{2}} \right]=\left[ {{a}_{11}},{{a}_{22}},......{{a}_{nn}} \right]......(3)$
We can write equation (3) as:
${{\left( {{a}_{11}} \right)}^{2}}={{a}_{11}};{{\left( {{a}_{22}} \right)}^{2}}={{a}_{22}};......;{{\left( {{a}_{nn}} \right)}^{2}}={{a}_{nn}}......(4)$

Simplifying the equation (4), we get:
${{\left( {{a}_{ii}} \right)}^{2}}={{a}_{ii}}$, for all elements i = 1,2,3……n

Now we can write equation (4) as:
${{\left( {{a}_{ii}} \right)}^{2}}-{{a}_{ii}}=0$
$\Rightarrow {{a}_{ii}}\left( {{a}_{ii}}-1 \right)=0$

Therefore, we get:
${{a}_{ii}}=0\text{ or }{{a}_{ii}}=1$, for all elements i = 1,2,3……n

For each element of diagonal, we have two choices, i.e. 0 or 1
Therefore, total values of matrix A are: $2\times 2\times 2\times .......n\text{ }times$
i.e. ${{2}^{n}}$
So, the correct answer is “Option C”.

Note: Since the diagonal matrix contains zero value of most of its elements, don’t choose option (a) directly.
Also, as given in the question: ${{A}^{2}}=A$, there may be a possibility of getting A = 1, which neglects the other choice. So be careful with these two options given.