Sample 2026-27 Class 12 Maths Q4 Solution: Symmetric matrix: value of √(6a + 2b + 3c)

The question

If $A = \begin{bmatrix} 2 & 1 & -2 \\ 3a & 0 & 4 \\ b & 2c & 1 \end{bmatrix}$ is a symmetric matrix, then value of $\sqrt{6a + 2b + 3c}$ is:

(A) $-2$

(B) $2$

(C) $\pm 2$

(D) $4$

Exam answer

$A$ is symmetric, so $A' = A$ ⁠, that is $a_{ij} = a_{ji}$ ⁠. Comparing $a_{21}$ with $a_{12}$ ⁠, $a_{31}$ with $a_{13}$ and $a_{32}$ with $a_{23}$ ⁠:

Why?

The transpose $A'$ turns rows into columns. A symmetric matrix equals its transpose, so each entry equals its mirror image across the leading diagonal: the entry in row $i$ ⁠, column $j$ equals the entry in row $j$ ⁠, column $i$ ⁠.

3a = 1,\quad b = -2,\quad 2c = 4

Why?

$3a$ (row 2, column 1) mirrors $1$ (row 1, column 2). $b$ (row 3, column 1) mirrors $-2$ (row 1, column 3). $2c$ (row 3, column 2) mirrors $4$ (row 2, column 3).

So $a = \frac{1}{3}$ ⁠, $b = -2$ and $c = 2$ ⁠.

\begin{aligned} 3a = 1 &\Rightarrow a = \frac{1}{3} \\ 2c = 4 &\Rightarrow c = \frac{4}{2} = 2 \end{aligned}

\begin{aligned} &6a + 2b + 3c \\ &= 6\left(\tfrac{1}{3}\right) + 2(-2) + 3(2) \\ &= 2 - 4 + 6 = 4 \end{aligned}

Why?

Work out each product, then add two at a time:

\begin{aligned} 6 \times \frac{1}{3} &= \frac{6}{3} = 2 \\ 2 \times (-2) &= -4 \\ 3 \times 2 &= 6 \\[4pt] 2 - 4 &= -2 \\ -2 + 6 &= 4 \end{aligned}

\begin{aligned} \sqrt{6a + 2b + 3c} &= \sqrt{2 - 4 + 6} \\ &= \sqrt{4} = 2 \end{aligned}

Why?

The symbol $\sqrt{\ }$ always means the non-negative square root, so $\sqrt{4} = 2$ only.

Watch out: $\pm 2$ in (C) are the two solutions of $y^2 = 4$ ⁠, which is a different question.

Answer: (B) $2$

Where this comes from

Chapter 3, Matrices: Symmetric matrices.

Question type: Transpose

How to spot it: click to see

Type 8. Transpose: symmetric or skew symmetric?

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

Here: $A$ is symmetric, so each entry equals its mirror: $3a = 1$ ⁠, $b = -2$ and $2c = 4$ ⁠. Then $\sqrt{6a + 2b + 3c} = \sqrt4 = 2$ ⁠.

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Q4: Symmetric matrix: value of 6a+2b+3c\sqrt{6a + 2b + 3c}6a+2b+3c​

Q4: Symmetric matrix: value of $\sqrt{6a + 2b + 3c}$