Let $A=\left(\begin{array}{cc}m & n \\ p & q\end{array}\right), d=|A| \neq 0, |A-d(\operatorname{Adj} A)|=0$. Then

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Matrices

Let

A = (m p n q), d = ∣ A ∣ \neq = 0, ∣ A - d (Adj A) ∣ = 0

. Then

Answer: (A)

With

A = [m p n q]

d = ∣ A ∣ = m q - n p \neq = 0

, the adjugate is

adj A = [q - p - n m]

, so

A - d (adj A) = [m - d q p (1 + d) n (1 + d) q - d m] .

Its determinant is

(m - d q) (q - d m) - n p (1 + d)^{2}

. Since

(m - d q) (q - d m) = m q (1 + d^{2}) - d (m^{2} + q^{2})

and

n p (1 + d)^{2} = n p (1 + d^{2}) + 2 d n p

, the determinant equals

(m q - n p) (1 + d^{2}) - d (m^{2} + q^{2}) - 2 d n p = d (1 + d^{2}) - d (m^{2} + q^{2}) - 2 d n p

. Dividing by

d \neq = 0

and using

2 n p = 2 m q - 2 d

gives

1 + d^{2} = (m^{2} + q^{2} + 2 m q) - 2 d = (m + q)^{2} - 2 d

, which rearranges to

(1 + d)^{2} = (m + q)^{2} .

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option