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MathMedMain 2026 • 2nd Apr - Shift 1

Matrix Equations and Determinant of Adjoint Matrix

Let A=[121α]A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix} and B=[33β2]B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}. If A2−4A+I=OA^2 - 4A + I = O and B2−5B−6I=OB^2 - 5B - 6I = O, then among the two statements : (S1): [(B−A)(B+A)]T=[1315710][(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix} and (S2): det⁡(adj(A+B))=−5\det(\text{adj}(A+B)) = -5,

Options

A

only (S1) is correct

B

only (S2) is correct

Correct
C

both (S1) and (S2) are correct

D

both (S1) and (S2) are wrong

Topics & Concepts

MatricesMatricesDeterminants

Step-by-Step Solution

Given the matrices: A=[121α]andB=[33β2]A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}

By the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation. For a 2×22 \times 2 matrix MM, the characteristic equation is given by: M2−tr(M)M+det⁡(M)I=OM^2 - \text{tr}(M)M + \det(M)I = O

For matrix AA: tr(A)=1+α,det⁡(A)=α−2\text{tr}(A) = 1 + \alpha, \quad \det(A) = \alpha - 2 So, the characteristic equation of AA is: A2−(1+α)A+(α−2)I=OA^2 - (1 + \alpha)A + (\alpha - 2)I = O

Comparing this with the given equation A2−4A+I=OA^2 - 4A + I = O: 1+α=4  ⟹  α=31 + \alpha = 4 \implies \alpha = 3 det⁡(A)=3−2=1\det(A) = 3 - 2 = 1

For matrix BB: tr(B)=3+2=5,det⁡(B)=6−3β\text{tr}(B) = 3 + 2 = 5, \quad \det(B) = 6 - 3\beta So, the characteristic equation of BB is: B2−5B+(6−3β)I=OB^2 - 5B + (6 - 3\beta)I = O

Comparing this with the given equation B2−5B−6I=OB^2 - 5B - 6I = O: 6−3β=−6  ⟹  3β=12  ⟹  β=46 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4

Thus, the matrices are: A=[1213]andB=[3342]A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}

Evaluation of Statement (S1):

Calculate B−AB - A and B+AB + A: B−A=[3−13−24−12−3]=[213−1]B - A = \begin{bmatrix} 3-1 & 3-2 \\ 4-1 & 2-3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}

B+A=[3+13+24+12+3]=[4555]B + A = \begin{bmatrix} 3+1 & 3+2 \\ 4+1 & 2+3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}

Now, calculate the product (B−A)(B+A)(B - A)(B + A): (B−A)(B+A)=[213−1][4555]=[2(4)+1(5)2(5)+1(5)3(4)+(−1)(5)3(5)+(−1)(5)]=[1315710](B - A)(B + A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 2(4) + 1(5) & 2(5) + 1(5) \\ 3(4) + (-1)(5) & 3(5) + (-1)(5) \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}

Taking the transpose: [(B−A)(B+A)]T=[1371510][(B - A)(B + A)]^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}

Since [(B−A)(B+A)]T≠[1315710][(B - A)(B + A)]^T \neq \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}, Statement (S1) is INCORRECT.

Evaluation of Statement (S2):

For any n×nn \times n matrix CC, the determinant of its adjoint is given by: det⁡(adj(C))=(det⁡(C))n−1\det(\text{adj}(C)) = (\det(C))^{n-1}

For n=2n = 2: det⁡(adj(A+B))=det⁡(A+B)\det(\text{adj}(A + B)) = \det(A + B)

We know that: A+B=[4555]A + B = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}

Now, compute det⁡(A+B)\det(A + B): det⁡(A+B)=(4)(5)−(5)(5)=20−25=−5\det(A + B) = (4)(5) - (5)(5) = 20 - 25 = -5

Thus, det⁡(adj(A+B))=−5\det(\text{adj}(A + B)) = -5. Statement (S2) is CORRECT.

Conclusion:

Only statement (S2) is correct.

Correct Option: B

Matrix Equations and Determinant of Adjoint Matrix | Mathematics PYQ Solution - JEE Challenger