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) 0
b) 1
c) $ {{2}^{n}} $
d) \[{{2}^{n-1}}\]

Answer 1

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.

Complete step-by-step answer:
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.