If A is a diagonal matrix of order n such that $A^{2} = A$ then the number of possible values of matrix A is $2^{n}$ . Whether it’s true or false.

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True

False

answer is A.

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Detailed Solution

A = [a_{11} \dots 0 ⋮ ⋱ ⋮ 0 \dots a_{nn}]

A^{2} = [a_{11} \dots 0 ⋮ ⋱ ⋮ 0 \dots a_{nn}] [a_{11} \dots 0 ⋮ ⋱ ⋮ 0 \dots a_{nn}]

A^{2} = [{{(a}_{11})}^{2} \dots 0 ⋮ ⋱ ⋮ 0 \dots {{(a}_{nn})}^{2}]

Given that

A^{2} = A

[{{(a}_{11})}^{2} \dots 0 ⋮ ⋱ ⋮ 0 \dots {{(a}_{nn})}^{2}] = [a_{11} \dots 0 ⋮ ⋱ ⋮ 0 \dots a_{nn}]

If two matrices of the same order are equal then their corresponding elements are also equal.

{(a_{11})}^{2} = a_{11} {(a_{22})}^{2} = a_{22}

……….

{(a_{nn})}^{2} = a_{nn}

{∴ (a_{ii})}^{2} = a_{ii}

{∴ (a_{ii})}^{2} - a_{ii} = 0

a_{ii} (a_{ii} - 1) = 0

a_{ii} = 0 or a_{ii} = 1

For each element of diagonal, we have two choices, i.e. 0 or 1
Therefore, total values of matrix A are: 2×2×2×....... n times
i.e.

2^{n}

Hence, the given statement is true.

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