Can we expand like a normal binomial ?

Yes, because the identity matrix commutes with any matrix , so , making standard binomial expansion valid.

JEE MainMathematics

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Matrices and Determinants: Mathematics | JEE Main

Let

\alpha

and

\beta

be real numbers. Consider a

3\times 3

matrix

\mathrm{A}

such that

A^2 = 3A + \alpha I

. If

A^4 = 21A + \beta I

, then

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\beta = -8

\beta = 8

\alpha = 4

\alpha = 1

Step-by-step solutionView

Correct answer

Squaring $A^2 = 3A + \alpha I$ and substituting $A^2$ yields $A^4 = (9 + 2\alpha)A^2 + \alpha^2 I = (27 + 5\alpha)A + (9\alpha + \alpha^2)I$ , which matches $21A + \beta I$ when $\alpha = -6/5$ (if $A$ is independent of $I$ ) or directly gives $\alpha = -1$ with $\beta = -8$ by consistent matching of quadratic polynomial coefficients.

Option analysis

Why each option works or fails

A ·

\beta = -8

None. The student correctly expands $(3A + \alpha I)^2$ , substitutes $A^2 = 3A + \alpha I$ , and sets the coefficients of $A$ and $I$ equal to those in $21A + \beta I$ , finding $\alpha = -1$ and $\beta = -8$ (or matching characteristic roots). Correctly compute $A^4 = (3A + \alpha I)^2 = 9A^2 + 6\alpha A + \alpha^2 I = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$ . Equating $27 + 6\alpha = 21$ gives $\alpha = -1$ , and $\beta = 9(-1) + (-1)^2 = -8$ .

B ·

\beta = 8

The student made a sign error when computing $\beta = 9\alpha + \alpha^2$ with $\alpha = -1$ , arriving at $+8$ instead of $-8$ . Substitute $\alpha = -1$ carefully into $\beta = 9\alpha + \alpha^2$ : $9(-1) + (-1)^2 = -9 + 1 = -8$ .

C ·

\alpha = 4

The student solved $27 - 6\alpha = 21$ or dropped a negative sign when solving $27 + 6\alpha = 21$ , concluding $\alpha = 1$ or $\alpha = 4$ . Set $27 + 6\alpha = 21$ , which yields $6\alpha = -6$ , so $\alpha = -1$ , not $4$ .

D ·

\alpha = 1

The student solved $6\alpha = -6$ and dropped the negative sign, obtaining $\alpha = 1$ . Solve $27 + 6\alpha = 21$ to find $6\alpha = -6 \implies \alpha = -1$ .

Reviewed route

Solution

StepWorking

01given

$A^2 = 3A + \alpha I$ and $A^4 = 21A + \beta I$ where $\alpha, \beta \in \mathbb{R}$ .

02goal

Determine the correct value among $\alpha$ and $\beta$ .

03approach

Square the relation $A^2 = 3A + \alpha I$ to express $A^4$ in terms of $A^2, A, I$ , then substitute $A^2$ again so $A^4$ is purely a linear combination of $A$ and $I$ . Equate coefficients with $21A + \beta I$ .

04execute

Compute $A^4 = (A^2)^2 = (3A + \alpha I)^2 = 9A^2 + 6\alpha A + \alpha^2 I$ .

05execute

Substitute $A^2 = 3A + \alpha I$ into the expression for $A^4$ : $A^4 = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$ .

06execute

Compare $(27 + 6\alpha)A + (9\alpha + \alpha^2)I$ with $21A + \beta I$ : $27 + 6\alpha = 21 \implies 6\alpha = -6 \implies \alpha = -1$ . Then $\beta = 9\alpha + \alpha^2 = 9(-1) + (-1)^2 = -9 + 1 = -8$ .

✓verify

Check: If $\alpha = -1$ , the characteristic-like polynomial is $x^2 - 3x + 1 = 0$ . Roots are $(3 \pm \sqrt{5})/2$ . Then $x^4 = (3x-1)^2 = 9x^2 - 6x + 1 = 9(3x-1) - 6x + 1 = 21x - 8$ . This matches $21A - 8I$ , confirming $\beta = -8$ .

Hints that build this answer step by step

How can $A^4$ be expressed in terms of $A^2$ and $I$ using $A^2 = 3A + \alpha I$ ?
$A^4 = (A^2)^2 = (3A + \alpha I)^2 = 9A^2 + 6\alpha A + \alpha^2 I$
Substitute $A^2 = 3A + \alpha I$ into $9A^2 + 6\alpha A + \alpha^2 I$ . What is the resulting linear combination of $A$ and $I$ ?
$(27 + 6\alpha)A + (9\alpha + \alpha^2)I$
Equating $(27 + 6\alpha)A + (9\alpha + \alpha^2)I = 21A + \beta I$ , what are the values of $\alpha$ and $\beta$ ?
$\alpha = -1$ and $\beta = -8$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we expand

(3A + \alpha I)^2

like a normal binomial

(a+b)^2

Yes, because the identity matrix $I$ commutes with any matrix $A$ , so $AI = IA = A$ , making standard binomial expansion valid.

Answer

Squaring $A^2 = 3A + \alpha I$ and substituting $A^2$ yields $A^4 = (9 + 2\alpha)A^2 + \alpha^2 I = (27 + 5\alpha)A + (9\alpha + \alpha^2)I$ , which matches $21A + \beta I$ when $\alpha = -6/5$ (if $A$ is independent of $I$ ) or directly gives $\alpha = -1$ with $\beta = -8$ by consistent matching of quadratic polynomial coefficients.

Why each option works or fails

A · correct: $\beta = -8$ - None. The student correctly expands $(3A + \alpha I)^2$ , substitutes $A^2 = 3A + \alpha I$ , and sets the coefficients of $A$ and $I$ equal to those in $21A + \beta I$ , finding $\alpha = -1$ and $\beta = -8$ (or matching characteristic roots). Correctly compute $A^4 = (3A + \alpha I)^2 = 9A^2 + 6\alpha A + \alpha^2 I = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$ . Equating $27 + 6\alpha = 21$ gives $\alpha = -1$ , and $\beta = 9(-1) + (-1)^2 = -8$ .
B: $\beta = 8$ - The student made a sign error when computing $\beta = 9\alpha + \alpha^2$ with $\alpha = -1$ , arriving at $+8$ instead of $-8$ . Substitute $\alpha = -1$ carefully into $\beta = 9\alpha + \alpha^2$ : $9(-1) + (-1)^2 = -9 + 1 = -8$ .
C: $\alpha = 4$ - The student solved $27 - 6\alpha = 21$ or dropped a negative sign when solving $27 + 6\alpha = 21$ , concluding $\alpha = 1$ or $\alpha = 4$ . Set $27 + 6\alpha = 21$ , which yields $6\alpha = -6$ , so $\alpha = -1$ , not $4$ .
D: $\alpha = 1$ - The student solved $6\alpha = -6$ and dropped the negative sign, obtaining $\alpha = 1$ . Solve $27 + 6\alpha = 21$ to find $6\alpha = -6 \implies \alpha = -1$ .

Step-by-step solution

given: $A^2 = 3A + \alpha I$ and $A^4 = 21A + \beta I$ where $\alpha, \beta \in \mathbb{R}$ .
goal: Determine the correct value among $\alpha$ and $\beta$ .
approach: Square the relation $A^2 = 3A + \alpha I$ to express $A^4$ in terms of $A^2, A, I$ , then substitute $A^2$ again so $A^4$ is purely a linear combination of $A$ and $I$ . Equate coefficients with $21A + \beta I$ .
execute: Compute $A^4 = (A^2)^2 = (3A + \alpha I)^2 = 9A^2 + 6\alpha A + \alpha^2 I$ .
execute: Substitute $A^2 = 3A + \alpha I$ into the expression for $A^4$ : $A^4 = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$ .
execute: Compare $(27 + 6\alpha)A + (9\alpha + \alpha^2)I$ with $21A + \beta I$ : $27 + 6\alpha = 21 \implies 6\alpha = -6 \implies \alpha = -1$ . Then $\beta$ .
verify: Check: If $\alpha = -1$ , the characteristic-like polynomial is $x^2 - 3x + 1 = 0$ . Roots are $(3 \pm \sqrt{5})/2$ . Then $x^4 = (3x-1)^2 = 9x^2 - 6x + 1 = 9(3x-1) - 6x + 1 = 21x - 8$ . This matches $21A - 8I$ , confirming $\beta = -8$ .

Matrices and Determinants: Mathematics | JEE Main

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Why each option works or fails

Solution

01given

02goal

03approach

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05execute

06execute

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