JEE Matrices Trick: Count Singular 2×2 Matrices in 59 Seconds! 🔥

 

❓ Question

The number of singular matrices of order 2, whose elements are taken from the set

$$\{2,3,6,9\},$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

A 2×2 matrix is singular if and only if its determinant is zero.
So the entire question reduces to counting how many choices make the determinant zero.

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🔹 Step 1 — Write the general 2×2 matrix

Let the matrix be:

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$$

where

$$a,b,c,d\in \{2,3,6,9\}\mathrm{.}$$

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🔹 Step 2 — Condition for singular matrix

A matrix is singular iff:

$$\mathrm{\mid }A\mathrm{\mid }=ad-bc=0$$

So we need:

$$ad=bc$$

👉 This is the ONLY condition to check.

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🔹 Step 3 — Key observation (MOST IMPORTANT 🔥)**

The set:

$$\{2,3,6,9\}$$

has a multiplicative structure:

$$2\times 3=6,3\times 3=9,2\times 6=12(\text{not allowed}),3\times 6=18(\text{not allowed})$$

So equality $ad = bc$ happens only when the ratios match:

$$\frac{a}{b}=\frac{c}{d}$$

👉 That means:

• Either rows are proportional

• Or columns are proportional

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🔹 Step 4 — Case-wise counting

🔸 Case 1: First row = Second row

That is:

$$(a,b)=(c,d)$$

Number of choices for $(a,b)$:

$$4\times 4=16$$

So 16 singular matrices from this case.

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🔸 Case 2: Second row is a multiple of first row (but not equal)**

Possible ratios within the set:

• Multiply by 3

• Multiply by 1/3

Valid proportional pairs from the set:

$$(2,3)\leftrightarrow (6,9)$$

So possible row pairs:

$$(2,3),(6,9)\text{or}(6,9),(2,3)$$

That gives 2 choices.

Similarly for columns:

$$(b,d)=(2,3),(6,9)\text{ or vice versa}$$

Counting carefully gives 8 more matrices.

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🔹 Step 5 — Total singular matrices

Adding all valid cases:

$$16+8=24\mathrm{<br>}$$

✅ Final Answer

$$\boxed{24}$$