Question

If A is a non-diagonal involutory matrix then
[a] A-I=O
[b] A+I = O
[c] A-I is non-zero singular
[d] none of these.

Answer 1

Hint:Use the definition of an involutory matrix is a square matrix which is inverse of itself. Hence a matrix is an involutory matrix if and only if it follows involution i.e. ${{\text{A}}^{2}}=\text{I}$. Use the fact that if A is a square matrix the $\text{AI=IA=A}$. Use the property of the distribution of matrix multiplication over matrix addition. Use the property that if A and B are square matrices, the $\det \left( \text{AB} \right)=\det \left( \text{A} \right)\det \left( \text{B} \right)$

Complete step-by-step answer:
We have A is an involutory matrix.
Hence ${{\text{A}}^{2}}=\text{I}$
Subtracting I from both sides, we get
${{\text{A}}^{2}}-\text{I=I-I=O}$
Adding and subtracting IA on LHS, we get
${{\text{A}}^{2}}-\text{IA+IA-I=O}$
Taking A common from the first two terms and I common from the last two terms, we have
$\text{A}\left( \text{A-I} \right)+\text{I}\left( \text{A-I} \right)=\text{O}$
Taking A-I common, we get
$\left( \text{A-I} \right)\left( \text{A+I} \right)=\text{O}$
Taking determinant on both sides, we get
$\det \left( \left( \text{A-I} \right)\left( \text{A+I} \right) \right)=\det \left( \text{O} \right)=0$
We know that det(AB) = det(A) det(B) if A and B are square matrices. Hence, we have
$\det \left( \text{A-I} \right)\det \left( \text{A+I} \right)=0$
Using zero product property, we get
$\det \left( \text{A-I} \right)=0$ or $\det \left( \text{A+I} \right)=0$
Hence either A-I is singular or A+I is singular.
Also since A is non-diagonal , we have $\text{A-I}\ne \text{O}$ and $\text{A+I}\ne \text{O}$.
This is because if A-I = O, then A = I which is diagonal and if A+I = O, then A = -I which is a diagonal matrix. But since A is non-diagonal we have that a diagonal matrix equals a non-diagonal matrix, which is a contradiction. Hence we have $\text{A-I}\ne \text{O}$ and $\text{A+I}\ne \text{O}$.
Hence option [c] is correct.

Note: Students usually make a mistake by saying that we factorise ${{\text{A}}^{2}}-{{\text{I}}^{2}}$ as $\left( \text{A-I} \right)\left( \text{A+I} \right)$ using the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, which is not correct since matrix multiplication is not commutative.
Hence in general $\left( \text{A}-\text{B} \right)\left( \text{A+B} \right)\ne {{\text{A}}^{2}}-{{\text{B}}^{2}}$.The above equation holds true for matrices which commute.
Since I commutes with every square matrix, we have ${{\text{A}}^{2}}-\text{I=}\left( \text{A}-\text{I} \right)\left( \text{A+I} \right)$.