Let $\mathrm{S}=\left\{\mathrm{A}=\left(\begin{array}{lll}0 & 1 & \mathrm{c} \\ 1 & \mathrm{a} & \mathrm{d} \\ 1 & \mathrm{~b} & \mathrm{e}\end{array}\right): \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e} \in\{0,1\}\right.$ and $\left.|\mathrm{A}| \in\{-1,1\}\right\}$, where $|\mathrm{A}|$ denotes the determinant of A. Then the number of elements in S is .........

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Determinants

Let

S = ⎩ ⎨ ⎧ A = 011 1 a b c d e : a, b, c, d, e \in {0, 1}

and

∣ A ∣ \in {- 1, 1}}

, where

∣ A ∣

denotes the determinant of A. Then the number of elements in S is .........

Answer: 16

The determinant is:

∣ A ∣ = 0 (a e - b d) - 1 (e - d) + c (b - a) = c (b - a) + (d - e)

We need

∣ A ∣ \in {- 1, 1}

with

a, b, c, d, e \in {0, 1}

.

**Case I:

c = 0

**

Then

∣ A ∣ = d - e

. For

∣ A ∣ = \pm 1

$d = 1, e = 0$ : $∣ A ∣ = 1$ . $a, b$ can be any of ${0, 1}$ , giving 4 matrices.
$d = 0, e = 1$ : $∣ A ∣ = - 1$ . $a, b$ can be any of ${0, 1}$ , giving 4 matrices.

Total: 8 matrices.

**Case II:

c = 1

**

Then

∣ A ∣ = (b - a) + (d - e)

. For

∣ A ∣ = \pm 1

$b = 1, a = 0$ : $b - a = 1$ . Need $d - e = 0$ , so $d = e$ . Two choices: $(d, e) \in {(0, 0), (1, 1)}$ . 2 matrices.
$b = 0, a = 1$ : $b - a = - 1$ . Need $d - e = 0$ , so $d = e$ . 2 matrices.
$b = 0, a = 0$ : $b - a = 0$ . Need $d - e = \pm 1$ . Two choices: $(d, e) \in {(1, 0), (0, 1)}$ . 2 matrices.
$b = 1, a = 1$ : $b - a = 0$ . Need $d - e = \pm 1$ . 2 matrices.

Total: 8 matrices.

Grand total: 16 matrices.

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Integer answer