The value (/s) of \( x \) satisfying the matrix equation:
\( \displaystyle x\begin{bmatrix}2x & 2\\3 & x\end{bmatrix}+2\begin{bmatrix}8 & 5x\\4 & 4x\end{bmatrix}=2\begin{bmatrix}(x^2+8) & 24\\10 & 6x\end{bmatrix} \)
is/are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( x=4 \) only
Multiplying out both sides gives
\( \displaystyle \begin{bmatrix}2x^2+16 & 12x\\3x+8 & x^2+8x\end{bmatrix}=\begin{bmatrix}2x^2+16 & 48\\20 & 12x\end{bmatrix} \)
Two matrices are equal only when every corresponding entry is equal:
\( \displaystyle (1,1):\ 2x^2+16=2x^2+16,\ \text{true for all } x \)
\( \displaystyle (1,2):\ 12x=48\ \Rightarrow\ x=4 \)
\( \displaystyle (2,1):\ 3x+8=20\ \Rightarrow\ x=4 \)
\( \displaystyle (2,2):\ x^2+8x=12x \)
\( \displaystyle x(x-4)=0\ \Rightarrow\ x=0 \text{ or } 4 \)
The value common to all four equations is x = 4 only.
Why the other options are incorrect
Option (1) \( x=0 \) and \( x=4 \) — Incorrect. x = 0 satisfies only the (2, 2) entry; with x = 0 the (1, 2) entries are 0 and 48.
Option (2) \( x=0 \) only — Incorrect. x = 0 fails the (1, 2) and (2, 1) entries.
Option (4) \( x=0 \) and \( x=-2 \) — Incorrect. Neither x = 0 nor x = −2 satisfies 12x = 48.