Let \(M=\left\{A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right): a, b, c, d \in\{ \pm 3, \pm 2, \pm 1,0\}\right\}\). Define \(f: M \rightarrow Z\), as \(f(A)=\operatorname{det}(A), \quad\) for all \(A \in M\), where \(Z\) is set of all integers. Then the number of \(A \in M\) such that \[ f(A)=15 \text { is } \] equal to

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Determinants

Let

M = {A = (a c b d) : a, b, c, d \in {\pm 3, \pm 2, \pm 1, 0}}

. Define

f : M \to Z

, as

f (A) = det (A),

for all

A \in M

, where

Z

is set of all integers. Then the number of

A \in M

such that

f (A) = 15 is

equal to

Answer: 16

Every entry lies in

{- 3, - 2, - 1, 0, 1, 2, 3}

, so each product of two entries lies between

- 9

and

9

. From

a d - b c = 15

with

b c \geq - 9

comes

a d \geq 6

, and with

a d \leq 9

comes

b c \leq - 6

.
Of the values

6, 7, 8, 9

, only

6

and

9

are products of two members of the set, so two cases remain.
If

a d = 6

then

b c = - 9

: the ordered pairs

(a, d)

are

(2, 3), (3, 2), (- 2, - 3), (- 3, - 2)

and the ordered pairs

(b, c)

are

(3, - 3), (- 3, 3)

, giving

4 \times 2 = 8

matrices.
If

a d = 9

then

b c = - 6

: the pairs

(a, d)

are

(3, 3), (- 3, - 3)

and the pairs

(b, c)

are

(2, - 3), (- 2, 3), (3, - 2), (- 3, 2)

, giving

2 \times 4 = 8

matrices.
The number of matrices in

M

with determinant

15

16

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer