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Question: If A is a non-diagonal involutory matrix then [a] A-I=O [b] A+I = O [c] A-I is non-zero singu...

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.

Explanation

Solution

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. A2=I{{\text{A}}^{2}}=\text{I}. Use the fact that if A is a square matrix the AI=IA=A\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⁡(AB)=det⁡(A)det⁡(B)\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 A2=I{{\text{A}}^{2}}=\text{I}
Subtracting I from both sides, we get
A2−I=I-I=O{{\text{A}}^{2}}-\text{I=I-I=O}
Adding and subtracting IA on LHS, we get
A2−IA+IA-I=O{{\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
A(A-I)+I(A-I)=O\text{A}\left( \text{A-I} \right)+\text{I}\left( \text{A-I} \right)=\text{O}
Taking A-I common, we get
(A-I)(A+I)=O\left( \text{A-I} \right)\left( \text{A+I} \right)=\text{O}
Taking determinant on both sides, we get
det⁡((A-I)(A+I))=det⁡(O)=0\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⁡(A-I)det⁡(A+I)=0\det \left( \text{A-I} \right)\det \left( \text{A+I} \right)=0
Using zero product property, we get
det⁡(A-I)=0\det \left( \text{A-I} \right)=0 or det⁡(A+I)=0\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 A-I≠O\text{A-I}\ne \text{O} and A+I≠O\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 A-I≠O\text{A-I}\ne \text{O} and A+I≠O\text{A+I}\ne \text{O}.
Hence option [c] is correct.

Note: Students usually make a mistake by saying that we factorise A2−I2{{\text{A}}^{2}}-{{\text{I}}^{2}} as (A-I)(A+I)\left( \text{A-I} \right)\left( \text{A+I} \right) using the formula (a+b)(a−b)=a2−b2\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 (A−B)(A+B)≠A2−B2\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 A2−I=(A−I)(A+I){{\text{A}}^{2}}-\text{I=}\left( \text{A}-\text{I} \right)\left( \text{A+I} \right).