1 markMCQSection A

If $\Delta_1 = \begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix}$ and $\Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix}$ , then

Q: If $\Delta_1 = \begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix}$ and $\Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix}$, then

**Correct option: (B) $\Delta_2 = -2\Delta_1$** $\Delta_1 = 1\cdot 2\cdot 3 = 6$ (diagonal determinant). Δ₂, expanded along the third column, equals $6 \times (0\cdot 0 - 2\cdot 1) = 6 \times (-2) = -12$. Since $-12 = -2 \times 6$, we get $\Delta_2 = -2\Delta_1$.

Determinants

Properties of Determinants

Options

(A)

\Delta_1 = 2 \Delta_2

(B)

\Delta_2 = -2 \Delta_1

(C)

\Delta_1 = \Delta_2

(D)

\Delta_2 = -\Delta_1

Official Answer

Correct option: (B) $\Delta_2 = -2\Delta_1$

$\Delta_1 = 1\cdot 2\cdot 3 = 6$ (diagonal determinant). Δ₂, expanded along the third column, equals $6 \times (0\cdot 0 - 2\cdot 1) = 6 \times (-2) = -12$ . Since $-12 = -2 \times 6$ , we get $\Delta_2 = -2\Delta_1$ .

determinant expansiondiagonal matrix determinant3x3 determinantcofactor expansiondeterminant propertiesrow and column operations

Marking Scheme

11 mark: correct option (B) selected — requires computing both determinants correctly.

Hint

Evaluate each $3\times 3$ determinant by expansion (Δ₁ is diagonal, so multiply the diagonal entries directly), then compare the numeric values.

Quick Oral Answer

Δ₁ evaluates to 6 and Δ₂ evaluates to -12 by cofactor expansion, and since $-12 = -2 \times 6$ , the relation is $\Delta_2 = -2\Delta_1$ , option (B).

Analysis & Explanation

This requires direct computation of two 3×3 determinants and comparing the results.

Computing Δ₁

Δ₁ is a diagonal matrix determinant: $\begin{vmatrix}1 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{vmatrix} = 1 \times 2 \times 3 = 6$ .

Computing Δ₂

$\Delta_2 = \begin{vmatrix}0 & 2 & 0\\1 & 0 & 0\\0 & 0 & 6\end{vmatrix}$ . Expanding along the third column (only the (3,3) entry = 6 is non-zero):
$\Delta_2 = 6 \times \begin{vmatrix}0 & 2\\1 & 0\end{vmatrix} = 6 \times (0\times 0 - 2\times 1) = 6 \times (-2) = -12$ .

Comparing the results

$\Delta_1 = 6$ and $\Delta_2 = -12$ , so $\Delta_2 = -2 \times 6 = -2\Delta_1$ , confirming option (B).

Why other options are wrong

(A) $\Delta_1 = 2\Delta_2$ would need $6 = 2\times(-12) = -24$ , which is false.
(C) $\Delta_1 = \Delta_2$ would need $6 = -12$ , false.
(D) $\Delta_2 = -\Delta_1$ would need $-12 = -6$ , false.

Common Mistakes

1Forgetting the sign from row interchange effects when expanding Δ₂, leading to $+12$ instead of $-12$ .
2Expanding along the wrong row/column and making arithmetic errors instead of choosing the row/column with the most zeros for easier expansion.

Interesting Facts

Interchanging any two rows (or columns) of a determinant changes its sign — this is exactly why $\Delta_2$ , which is $\Delta_1$ with rows 1 and 2 swapped and column 2 scaled, comes out negative and scaled.

Determinants were originally developed by Japanese mathematician Seki Takakazu and German mathematician Gottfried Leibniz independently in the late 17th century, before matrix notation itself was formalized.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

How do you quickly evaluate a diagonal matrix's determinant?

For any diagonal matrix, the determinant equals the product of the diagonal entries, since all cofactor expansions along rows/columns with zero off-diagonal entries collapse to just multiplying the diagonal terms.

Why did swapping rows change the sign of the determinant here?

One of the fundamental properties of determinants is that interchanging any two rows (or columns) multiplies the determinant by $-1$ . $\Delta_2$ is related to $\Delta_1$ through such row/column manipulations combined with scaling, which is why the final value came out negative and scaled by a factor.