JEE MainMathematics

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Finding Coefficients in Matrix Polynomial Relations

Q: Can we expand (3A + α 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.

Let α and β be real numbers. Consider a 3× 3 matrix A such that A^2 = 3A + α I. If A^4 = 21A + β I, then

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Aβ = -8Bβ = 8Cα = 4Dα = 1

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A · β = -8

The student correctly expands $(3 A + α I)^{2}$ , substitutes $A^{2} = 3 A + α I$ , and sets the coefficients of $A$ and $I$ equal to those in $21 A + β I$ , finding $α = - 1$ and $β = - 8$ (or matching characteristic roots). Correctly compute $A^{4} = (3 A + α I)^{2} = 9 A^{2} + 6 α A + α^{2} I = 9 (3 A + α I) + 6 α A + α^{2} I = (27 + 6 α) A + (9 α + α^{2}) I$ . Equating $27 + 6 α = 21$ gives $α = - 1$ , and $β = 9 (- 1) + (- 1)^{2} = - 8$ .

B · β = 8

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

C · α = 4

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

D · α = 1

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

Reviewed route

Solution

StepWorking

01Given

$A^{2} = 3 A + α I$ and $A^{4} = 21 A + β I$ where $α, β \in R$ .

02Goal

Determine the correct value among $α$ and $β$ .

03Approach

Square the relation $A^{2} = 3 A + α I$ 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 $21 A + β I$ .

04Execute

Compute $A^{4} = (A^{2})^{2} = (3 A + α I)^{2} = 9 A^{2} + 6 α A + α^{2} I$ .

05Execute

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

06Execute

Compare $(27 + 6 α) A + (9 α + α^{2}) I$ with $21 A + β I$ : $27 + 6 α = 21 ⟹ 6 α = - 6 ⟹ α = - 1$ . Then $β = 9 α + α^{2} = 9 (- 1) + (- 1)^{2} = - 9 + 1 = - 8$ .

✓Verify

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

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we expand

(3 A + α I)^{2}

like a normal binomial

(a + b)^{2}

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

Answer

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

Why each option works or fails

A · correct: β = -8 - The student correctly expands $(3 A + α I)^{2}$ , substitutes $A^{2} = 3 A + α I$ , and sets the coefficients of $A$ and $I$ equal to those in $21 A + β I$ , finding $α = - 1$ and $β = - 8$ (or matching characteristic roots). Correctly compute $A^{4} = (3 A + α I)^{2} = 9 A^{2} + 6 α A + α^{2} I = 9 (3 A + α I) + 6 α A + α^{2} I = (27 + 6 α) A + (9 α + α^{2}) I$ . Equating $27 + 6 α = 21$ gives $α = - 1$ , and $β = 9 (- 1) + (- 1)^{2} = - 8$ .
B: β = 8 - The student made a sign error when computing $β = 9 α + α^{2}$ with $α = - 1$ , arriving at $+ 8$ instead of $- 8$ . Substitute $α = - 1$ carefully into $β = 9 α + α^{2}$ : $9 (- 1) + (- 1)^{2} = - 9 + 1 = - 8$ .
C: α = 4 - The student solved $27 - 6 α = 21$ or dropped a negative sign when solving $27 + 6 α = 21$ , concluding $α = 1$ or $α = 4$ . Set $27 + 6 α = 21$ , which yields $6 α = - 6$ , so $α = - 1$ , not $4$ .
D: α = 1 - The student solved $6 α = - 6$ and dropped the negative sign, obtaining $α = 1$ . Solve $27 + 6 α = 21$ to find $6 α = - 6 ⟹ α = - 1$ .

Step-by-step solution

Given: $A^{2} = 3 A + α I$ and $α = - 1$ where $α, β \in R$ .
Goal: Determine the correct value among $α$ and $β$ .
Approach: Square the relation $A^{2} = 3 A + α I$ 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 $21 A + β I$ .
Execute: Compute $A^{4} = (A^{2})^{2} = (3 A + α I)^{2} = 9 A^{2} + 6 α A + α^{2} I$ .
Execute: Substitute $A^{2} = 3 A + α I$ into the expression for $A^{4}$ : $A^{4} = 9 (3 A + α I) + 6 α A + α^{2} I = (27 + 6 α) A + (9 α + α^{2}) I$ .
Execute: Compare $(27 + 6 α) A + (9 α + α^{2}) I$ with $21 A + β I$ : $27 + 6 α = 21 ⟹ 6 α = - 6 ⟹ α = - 1$ . Then $β = 9 α + α^{2} = 9 (- 1) + (- 1)^{2} = - 9 + 1 = - 8$ .
Verify: Check: If $α = - 1$ , the characteristic-like polynomial is $x^{2} - 3 x + 1 = 0$ . Roots are $(3 \pm \sqrt{5}) / 2$ . Then $x^{4} = (3 x - 1)^{2} = 9 x^{2} - 6 x + 1 = 9 (3 x - 1) - 6 x + 1 = 21 x - 8$ . This matches $21 A - 8 I$ , confirming $β = - 8$ .

Finding Coefficients in Matrix Polynomial Relations

What feels right?

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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