Question

Let $A = \begin{bmatrix} 2 & x \\ 3 & 1 \end{bmatrix}, B = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 3x+2 & 10 \\ 14-x & 10 \end{bmatrix}$. If $B^T A^T = C$, then the value of $x$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Hint:

Always use $(AB)^T = B^T A^T$ to avoid transposing multiple matrices separately. Transpose the final result once to save time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Using the reversal law of transposes, $B^T A^T = (AB)^T$. Therefore, $(AB)^T = C$, which implies $AB = C^T$.

Step 2: Detailed Explanation:
1. Calculate product $AB$:
\[ AB = \begin{bmatrix} 2 & x \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 2(3)+x(1) & 2(2)+x(4) \\ 3(3)+1(1) & 3(2)+1(4) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 6+x & 4+4x \\ 10 & 10 \end{bmatrix} \]
2. Transpose $C$:
\[ C^T = \begin{bmatrix} 3x+2 & 14-x \\ 10 & 10 \end{bmatrix} \]
3. Equate elements:
From the first element $(1,1)$:
\[ 6+x = 3x+2 \]
\[ 2x = 4 \implies x = 2 \]
4. Verify with element $(1,2)$:
\[ 4+4(2) = 4+8 = 12 \]
\[ 14-2 = 12 \]
The equality holds for $x=2$.

Step 3: Final Answer:
The value of $x$ is 2.