The number of singular matrices of order 2 , whose elements are from the set \2,3,6,9\ , is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q46 with Solution

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Correct answer: 36

Let the matrix be $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ where each entry is chosen from the set $S=\{2,3,6,9\}.$

We need the number of singular matrices of order $2$ .

A $2\times 2$ matrix is singular if and only if its determinant is zero: $\det(A)=ad-bc=0$ so we need $ad=bc.$

1. Total approach

We count the number of ordered quadruples $(a,b,c,d)\in S^4$ such that $ad=bc.$

Instead of checking all $4^4=256$ matrices directly, we count by products.

2. Possible products from $S$

Take all ordered pairs $(x,y)\in S^2$ and compute $xy$ .

From $S=\{2,3,6,9\}$ , the possible products are:

$2\cdot 2=4$
$2\cdot 3=6$
$2\cdot 6=12$
$2\cdot 9=18$
$3\cdot 2=6$
$3\cdot 3=9$
$3\cdot 6=18$
$3\cdot 9=27$
$6\cdot 2=12$
$6\cdot 3=18$
$6\cdot 6=36$
$6\cdot 9=54$
$9\cdot 2=18$
$9\cdot 3=27$
$9\cdot 6=54$
$9\cdot 9=81$

Now count how many ordered pairs produce each product:

\text{Product }p & \#\{(x,y)\in S^2:xy=p\} \\\hline 4 & 1 \\ 6 & 2 \\ 9 & 1 \\ 12 & 2 \\ 18 & 4 \\ 27 & 2 \\ 36 & 1 \\ 54 & 2 \\ 81 & 1 \end{array}$$ Let this count be $f(p)$. ## 3. Count matrices with $ad=bc$ For a singular matrix, product $ad$ must equal product $bc$. For each product $p$, we can choose: - $(a,d)$ in $f(p)$ ways, - $(b,c)$ in $f(p)$ ways. So number of matrices corresponding to product $p$ is $$f(p)^2.$$ Hence total number of singular matrices is $$\sum_p f(p)^2.$$ Now compute: $$1^2+2^2+1^2+2^2+4^2+2^2+1^2+2^2+1^2$$ $$=1+4+1+4+16+4+1+4+1$$ $$=36.$$ ## 4. Final answer Therefore, the number of singular $2\times 2$ matrices is $$\boxed{36}.$$

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2. Possible products from

S

Take all ordered pairs

(x,y)\in S^2

and compute

xy

From

S=\{2,3,6,9\}

, the possible products are:

2\cdot 2=4

2\cdot 3=6

2\cdot 6=12

2\cdot 9=18

3\cdot 2=6

3\cdot 3=9

3\cdot 6=18

3\cdot 9=27

6\cdot 2=12

6\cdot 3=18

6\cdot 6=36

6\cdot 9=54

9\cdot 2=18

9\cdot 3=27

9\cdot 6=54

9\cdot 9=81

Now count how many ordered pairs produce each product:

\text{Product }p & \#\{(x,y)\in S^2:xy=p\} \\\hline 4 & 1 \\ 6 & 2 \\ 9 & 1 \\ 12 & 2 \\ 18 & 4 \\ 27 & 2 \\ 36 & 1 \\ 54 & 2 \\ 81 & 1 \end{array}$$ Let this count be $f(p)$. ## 3. Count matrices with $ad=bc$ For a singular matrix, product $ad$ must equal product $bc$. For each product $p$, we can choose: - $(a,d)$ in $f(p)$ ways, - $(b,c)$ in $f(p)$ ways. So number of matrices corresponding to product $p$ is $$f(p)^2.$$ Hence total number of singular matrices is $$\sum_p f(p)^2.$$ Now compute: $$1^2+2^2+1^2+2^2+4^2+2^2+1^2+2^2+1^2$$ $$=1+4+1+4+16+4+1+4+1$$ $$=36.$$ ## 4. Final answer Therefore, the number of singular $2\times 2$ matrices is $$\boxed{36}.$$

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