Let $\mathrm{M}=\left[\begin{array}{lll}0 & 1 & \mathrm{a} \\ 1 & 2 & 3 \\ 3 & \mathrm{~b} & 1\end{array}\right]$ and $\operatorname{adj} \mathrm{M}=\left[\begin{array}{ccc}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]$ where a and b are real numbers. Which of the following options is/are correct ?

(A) $\mathrm{a}+\mathrm{b}=3$; (C) $(\operatorname{adj} \mathrm{M})^{-1}+\mathrm{adjM}^{-1}=-\mathrm{M}$; (D) If $\mathrm{M}\left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$, then $\alpha-\beta+\gamma=3$

(B) $\operatorname{det}\left(\operatorname{adjM}^{2}\right)=81$

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Matrices

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Let

M = 013 12 b a 31

and

adj M = - 1 8 - 5 1 - 6 3 - 1 2 - 1

where a and b are real numbers. Which of the following options is/are correct ?

(A) $a + b = 3$ ✓ Correct
(B) $det (adjM^{2}) = 81$
(C) $(adj M)^{- 1} + adjM^{- 1} = - M$ ✓ Correct
(D)If $M α β γ = 123$ , then $α - β + γ = 3$ ✓ Correct

Answer: (A), (C), (D)

Solution

From

(adjM)_{11} = 2 - 3 b = - 1

, we get

b = 1

.
From

(adjM)_{22} = - 3 a = - 6

, we get

a = 2

.
So

a + b = 3

✓

det M = - 2

, so

det (adj M^{2}) = (det M)^{4} = 16

✗

Since

M^{- 1} = \frac{adjM}{det M}

, we have

adj M = - 2 M^{- 1}

.
Thus

(adj M)^{- 1} = \frac{- 1}{2} M

.

Also

adj (M^{- 1}) = \frac{1}{det M} M = \frac{- M}{2}

.
Hence

(adj M)^{- 1} + adj (M^{- 1}) = - M

✓

For the system

MX = b

X = M^{- 1} b = \frac{- adjM}{2} b = \frac{- 1}{2} - 2 2 - 2 = 1 - 1 1

(α, β, γ) = (1, - 1, 1)

and

α - β + γ = 3

✓

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Matrices

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