If $A$ is a square matrix such that $A^2 = I$, then $(A-I)^3 + (A+I)^3 - 7A$ is equal to (A) $A$ (B) $I - A$ (C) $I + A$ (D) $3A$

If A is a square matrix such that A² = I, then (A- – Class 12 Mathematics

NCERT Exemplar · Q62

A

Uttarakhand UbseMCQ· 1mImportance★★★★★

Appeared in past exams:GUJCET 2020· Set 07· 1mexact

79% · 143/182 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

You know how to evaluate a polynomial like $p(x) = 2x^2 - 3x + 5$ at a number: plug in $x$ , get a number out. Now plug in a square matrix $A$ instead. The variable becomes $A$ , and — crucially — the constant term becomes a multiple of the identity matrix $I$ , because you cannot add a bare number to a matrix.

The Definition

p(x) = a_n x^n + \cdots + a_1 x + a_0

Here $A^k$ is $k$ -fold matrix multiplication, $a_k A^k$ is scalar multiplication, and $a_0 I$ replaces the constant. The result is a square matrix of the same size as $A$ .

Note

There is no ambiguity from non-commutativity here: a polynomial only ever multiplies $A$ by itself, and $A$ always commutes with $A$ .

A Worked Example

Let $p(x) = x^2 - 4x + 3$ and $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ .

A^2 = \begin{pmatrix} 4 & 5 \\ 0 & 9 \end{pmatrix}, \quad -4A = \begin{pmatrix} -8 & -4 \\ 0 & -12 \end{pmatrix}, \quad 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}.

Adding term by term,

p(A) = \begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}.

A Shortcut for Diagonal Matrices

If $A = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$ , then $A^k = \begin{pmatrix} \lambda_1^k & 0 \\ 0 & \lambda_2^k \end{pmatrix}$ , so

p(A) = \begin{pmatrix} p(\lambda_1) & 0 \\ 0 & p(\lambda_2) \end{pmatrix}.

You simply evaluate $p$ at each diagonal entry. …

Start your 14-day free trial to unlock the full solution →

The key idea is that $A^2 = I$ makes $A$ an involutory matrix, so powers of $A$ simplify cyclically. Expanding the cubes and simplifying using $A^2 = I$ gives the result as $A$ . The final answer is option (A).

We are given that $A$ is a square matrix with $A^2 = I$ . This is the defining property of an involutory matrix — a matrix that is its own inverse. Because of this, any higher power of $A$ reduces to either $A$ or $I$ :

$A^3 = A^2 \cdot A = I \cdot A = A$ ,

$A^4 = A^2 \cdot A^2 = I \cdot I = I$ , and so on. This cyclic behaviour (period 2) is the engine that will simplify the expression.

The expression to evaluate is:

(A-I)^3 + (A+I)^3 - 7A

We could expand each cube using the binomial theorem, but we must remember that matrix multiplication is not commutative in general — however, here $A$ and $I$ commute (since $I$ commutes with every matrix), so we can safely expand as if they were numbers.

Let’s work through it step by step.

Expand $(A-I)^3$

(A-I)^3 = A^3 - 3A^2I + 3AI^2 - I^3

Since $I$ is the identity, $I^2 = I$ , $I^3 = I$ , and $AI = A$ , $A^2I = A^2$ . So:

(A-I)^3 = A^3 - 3A^2 + 3A - I

Expand $(A+I)^3$

(A+I)^3 = A^3 + 3A^2I + 3AI^2 + I^3 = A^3 + 3A^2 + 3A + I

Add the two expansions

(A-I)^3 + (A+I)^3 = (A^3 - 3A^2 + 3A - I) + (A^3 + 3A^2 + 3A + I)

The $-3A^2$ and $+3A^2$ cancel. The $-I$ and $+I$ cancel. We are left with:

= 2A^3 + 6A

Simplify $A^3$ using $A^2 = I$ Since $A^3 = A^2 \cdot A = I \cdot A = A$ , we get: …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ61 — If A and B are matrices of same order, then (AB' - BA') is a (A) skew …→Next questionQ63 — For any two matrices A and B, we have (A) AB = BA (B) AB ≠ BA (C) AB =…

Q.If $A$ is a square matrix such that $A^2 = I$ , then $(A-I)^3 + (A+I)^3 - 7A$ is equal to
(A) $A$
(B) $I - A$
(C) $I + A$
(D) $3A$

Concept understanding — Matrix Polynomial Evaluation

Matrix Polynomial Evaluation

The Definition

A Worked Example

A Shortcut for Diagonal Matrices

Method: Simplify a matrix polynomial using a given power identity

Steps

Common Mistakes

Q.If AAA is a square matrix such that A2=IA^2 = IA2=I, then (A−I)3+(A+I)3−7A(A-I)^3 + (A+I)^3 - 7A(A−I)3+(A+I)3−7A is equal to (A) AAA (B) I−AI - AI−A (C) I+AI + AI+A (D) 3A3A3A

🎓Unlock everything free for 14 days

Q.If $A$ is a square matrix such that $A^2 = I$ , then $(A-I)^3 + (A+I)^3 - 7A$ is equal to
(A) $A$
(B) $I - A$
(C) $I + A$
(D) $3A$

Unlock everything free for 14 days