Sample 2023-24 Class 12 Maths Q10 Solution: |A| for a 3 × 3 skew symmetric matrix in x

The question

The value of $|A|$ ⁠, if

A = \begin{bmatrix} 0 & 2x - 1 & \sqrt x \\ 1 - 2x & 0 & 2\sqrt x \\ -\sqrt x & -2\sqrt x & 0 \end{bmatrix},

where $x \in \mathbb{R}^+$ ⁠, is

(a) $(2x + 1)^2$

(b) $0$

(c) $(2x + 1)^3$

(d) $(2x - 1)^2$

Exam answer

Every diagonal entry is 0 and $a_{ji} = -a_{ij}$ for $i \ne j$ ⁠, so $A' = -A$ ⁠: $A$ is skew symmetric of order 3.

Why?

Check the pairs across the diagonal:

\begin{aligned} a_{21} &= 1 - 2x = -a_{12} \\ a_{31} &= -\sqrt x = -a_{13} \\ a_{32} &= -2\sqrt x = -a_{23} \end{aligned}

\begin{aligned} |A| &= |A'| = \lvert -A\rvert \\ &= (-1)^3|A| = -|A| \end{aligned}

Why?

Two standard results for a square matrix of order $n$ ⁠:

\begin{aligned} |A'| &= |A| \\ |kA| &= k^n|A| \end{aligned}

Here $k = -1$ and $n = 3$ ⁠, so the factor is $(-1)^3 = -1$ ⁠.

$\therefore\ 2|A| = 0$ ⁠, so $|A| = 0$ ⁠.

Why?

$|A| = -|A|$ means $|A| + |A| = 0$ ⁠.

\begin{aligned} |A| &= -|A| \\ |A| + |A| &= 0 \quad (\text{add } |A|) \\ 2|A| &= 0 \\ |A| &= \frac{0}{2} = 0 \end{aligned}

Watch out: this works only for odd order. For order 2, $(-1)^2 = 1$ and the same steps just give $|A| = |A|$ ⁠.

Options (a) and (c) are never 0 for $x > 0$ ⁠, and (d) is 0 only at $x = \frac{1}{2}$ ⁠.

Answer: (b) $0$

Where this comes from

Chapter 4, Determinants: Determinant of a skew symmetric matrix.

Question type: Transpose + One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$ (combines 2 types)

How to spot it: click to see

Ch 3, type 8. Transpose: symmetric or skew symmetric?

Spot it: "show that $A$ is symmetric", "show that $A$ is skew symmetric", or "find $A'$ ".

Here: First, compare entries across the diagonal: the diagonal is all 0, and $a_{21} = 1 - 2x = -a_{12}$ ⁠, and so on. So $A' = -A$ ⁠: $A$ is skew symmetric.

Ch 4, type 10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Here: Then use the cards with order 3: $|A| = |A'| = |-A| = (-1)^3|A| = -|A|$ ⁠, so $|A| = 0$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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