Let M = \ A = [ cases a & b c & d cases ]: a, b, c, d ∈ (± 3, ± 2, ± 1, 0)\ . Define f: M → Z, as f(A) = det (A), for all A ∈ M where Z is set of all integers. Then the number of A ∈ M such that f(A) = 15 is equal to .

Correct answer: 0 . |A| = ad - bc = 15 where a, b, c, d ∈ \ ± 3, ± 2, ± 1, 0\ Case I: ad = 9 and bc = - 6 For ad possible pairs are (3, 3), (- 3, - 3) . For bc possible pairs are (3, - 2), (- 3, 2), (- 2, 3), (2, - 3) So, total number of matrices in case I = 2 × 4 = 8 Case II: ad = 6 and bc = - 9 Similarly, total number of matrices in case II = 2 × 4 = 8 Hence, total number of matrices are 16 .

JEE Main 202125 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 25 July · Morning Shift · Mathematics · Matrices

Let M = \ A = [ cases a & b c & d cases ]: a, b, c, d ∈ (± 3, ± 2,...

Let

M = \left\{A = \left[

\begin{cases}a & b \\ c & d\end{cases}

\right]: a, b, c, d \in \left(\pm 3, \pm 2, \pm 1, 0\right)\right\}

. Define

f: M \to Z,

f\left(A\right) = det \left(A\right),

for all

A \in M

where

Z

is set of all integers. Then the number of

A \in M

such that

f\left(A\right) = 15

is equal to .

Correct answer

Step-by-step solution

. $\left|A\right| = ad - bc = 15$

where $a, b, c, d \in \left\{\pm 3, \pm 2, \pm 1, 0\right\}$

Case $I:$ $ad = 9 \text{ and } bc = - 6$

For $ad$ possible pairs are \left(3, 3\right), \left(- 3, - 3\right) . $</p><p>For$ bc $possible pairs are \left(3, - 2\right), \left(- 3, 2\right), \left(- 2, 3\right), \left(2, - 3\right)$

So, total number of matrices in case $I$ $= 2 \times 4 = 8$

Case $II:$ $ad = 6 \text{ and } bc = - 9$

Similarly, total number of matrices in case $II$ $= 2 \times 4 = 8$

Hence, total number of matrices are $16$ .

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