Let the determinant of a square matrix A of order m be \(\mathrm{m}-\mathrm{n}\), where m and n satisfy \(4 \mathrm{~m}+\mathrm{n}=22\) and \(17 \mathrm{~m}+4 \mathrm{n}=93\). If det \((\mathrm{n} \operatorname{adj}(\operatorname{adj}(\mathrm{mA})))= 3^{a} 5^{b} 6^{c}\), then \(\mathrm{a}+\mathrm{b}+\mathrm{c}\) is equal to:

Matrices — JEE Main · 2023 · 15 Apr 2023 · Shift 1 Question with Solution

Let the determinant of a square matrix A of order m be

m - n

, where m and n satisfy

4 m + n = 22

and

17 m + 4 n = 93

. If det

(n adj (adj (mA))) = 3^{a} 5^{b} 6^{c}

, then

a + b + c

is equal to:

Answer: (A)

Solve the system:

4 m + n = 22

17 m + 4 n = 93

→

m = 5, n = 2

.

So

∣ A ∣ = m - n = 3

, order of A is

m = 5

.

Using

∣ adj (B) ∣ = ∣ B ∣^{m - 1}

for

m \times m

matrix B:

∣ adj (adj (5 A)) ∣ = ∣5 A ∣^{(5 - 1)^{2}} = ∣5 A ∣^{16} = (5^{5} ∣ A ∣)^{16} = 5^{80} \cdot 3^{16}

.

Thus

∣2 adj (adj (5 A)) ∣ = 2^{5} \cdot 5^{80} \cdot 3^{16} = 3^{16} \cdot 5^{80} \cdot 2^{5}

.

Rewrite

2^{5} = 32 = 2 \cdot 16 = 2 \cdot 2^{4}

, but cleaner:

3^{16} \cdot 5^{80} \cdot 2^{5} = 3^{11} \cdot 5^{80} \cdot 6^{5}

(since

6^{5} = 2^{5} \cdot 3^{5}

, so

3^{16} = 3^{11} \cdot 3^{5}

).

Thus

a = 11, b = 80, c = 5

, and

a + b + c = 96

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option