JEE Challenger
More from Matrices
MathMedAdv 2019 • Paper 1

Properties and Inverse Relations of Matrix M and its Adjoint

Let the matrix MM and its adjoint be given by

M=[01a1233b1]andadj M=[−11−18−62−53−1]M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix} \quad \text{and} \quad \text{adj } M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}

where aa and bb are real numbers. Which of the following statements is/are correct?

Options

A

a+b=3a + b = 3

Correct
B

(adj M)−1+adj M−1=−M(\text{adj } M)^{-1} + \text{adj } M^{-1} = -M

Correct
C

det⁡(adj M2)=81\det(\text{adj } M^2) = 81

D

If M[αβγ]=[123]M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, then α−β+γ=3\alpha - \beta + \gamma = 3

Correct

Topics & Concepts

MatricesMatricesDeterminants

Step-by-Step Solution

To determine which statements are correct, let us analyze the given matrix MM and its adjoint adj M\text{adj } M:

M=[01a1233b1]andadj M=[−11−18−62−53−1]M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix} \quad \text{and} \quad \text{adj } M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}

Step 1: Finding aa, bb, and det⁡(M)\det(M)

We use the fundamental identity:

M⋅(adj M)=(det⁡M)I3=[det⁡M000det⁡M000det⁡M]M \cdot (\text{adj } M) = (\det M) I_3 = \begin{bmatrix} \det M & 0 & 0 \\ 0 & \det M & 0 \\ 0 & 0 & \det M \end{bmatrix}

Multiplying the first row of MM with the columns of adj M\text{adj } M:

  1. With column 2: 0(1)+1(−6)+a(3)=0  ⟹  −6+3a=0  ⟹  a=20(1) + 1(-6) + a(3) = 0 \implies -6 + 3a = 0 \implies a = 2

  2. With column 1: 0(−1)+1(8)+a(−5)=det⁡M  ⟹  8−5(2)=det⁡M  ⟹  det⁡M=−20(-1) + 1(8) + a(-5) = \det M \implies 8 - 5(2) = \det M \implies \det M = -2

Multiplying the third row of MM with the first column of adj M\text{adj } M: 3(−1)+b(8)+1(−5)=0  ⟹  −8+8b=0  ⟹  b=13(-1) + b(8) + 1(-5) = 0 \implies -8 + 8b = 0 \implies b = 1

Thus, we have: a=2,b=1,anddet⁡M=−2a = 2, \quad b = 1, \quad \text{and} \quad \det M = -2

Step 2: Evaluating the Options

Option (A): a+b=2+1=3a + b = 2 + 1 = 3 Hence, Option (A) is correct.

Option (B): For any invertible matrix MM of order 3×33 \times 3, we know that: adj M=(det⁡M)M−1  ⟹  (adj M)−1=1det⁡MM\text{adj } M = (\det M) M^{-1} \implies (\text{adj } M)^{-1} = \frac{1}{\det M} M Similarly, for the inverse matrix M−1M^{-1}: adj(M−1)=det⁡(M−1)(M−1)−1=1det⁡MM\text{adj}(M^{-1}) = \det(M^{-1}) (M^{-1})^{-1} = \frac{1}{\det M} M

Adding the two equations: (adj M)−1+adj(M−1)=1det⁡MM+1det⁡MM=2det⁡MM(\text{adj } M)^{-1} + \text{adj}(M^{-1}) = \frac{1}{\det M} M + \frac{1}{\det M} M = \frac{2}{\det M} M Since det⁡M=−2\det M = -2: (adj M)−1+adj(M−1)=2−2M=−M(\text{adj } M)^{-1} + \text{adj}(M^{-1}) = \frac{2}{-2} M = -M Hence, Option (B) is correct.

Option (C): For any 3×33 \times 3 matrix AA, the determinant of its adjoint is given by: det⁡(adj A)=(det⁡A)3−1=(det⁡A)2\det(\text{adj } A) = (\det A)^{3 - 1} = (\det A)^2

Applying this to A=M2A = M^2: det⁡(adj M2)=(det⁡(M2))2=((det⁡M)2)2=(det⁡M)4\det(\text{adj } M^2) = (\det(M^2))^2 = ((\det M)^2)^2 = (\det M)^4 Substituting det⁡M=−2\det M = -2: det⁡(adj M2)=(−2)4=16≠81\det(\text{adj } M^2) = (-2)^4 = 16 \neq 81 Hence, Option (C) is incorrect.

Option (D): Given the equation:

M[αβγ]=[123]M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}

Multiplying both sides by M−1=1det⁡Madj MM^{-1} = \frac{1}{\det M} \text{adj } M:

[αβγ]=1−2[−11−18−62−53−1][123]\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}

Computing the matrix-vector product:

[−11−18−62−53−1][123]=[−1(1)+1(2)−1(3)8(1)−6(2)+2(3)−5(1)+3(2)−1(3)]=[−22−2]\begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} -1(1) + 1(2) - 1(3) \\ 8(1) - 6(2) + 2(3) \\ -5(1) + 3(2) - 1(3) \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix}

Therefore:

[αβγ]=−12[−22−2]=[1−11]\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -2 \\ 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

Thus, α=1\alpha = 1, β=−1\beta = -1, and γ=1\gamma = 1. Evaluating the given expression: α−β+γ=1−(−1)+1=3\alpha - \beta + \gamma = 1 - (-1) + 1 = 3 Hence, Option (D) is correct.

Conclusion:

The correct options are A, B, and D.

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