If \( \begin{bmatrix}2x & 4\end{bmatrix}\begin{bmatrix}x\\-8\end{bmatrix}=\begin{bmatrix}0\end{bmatrix} \), then the value of x is/are
(A) −4
(B) 4
(C) 16
(D) −16
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
A 1 × 2 matrix times a 2 × 1 matrix is a 1 × 1 matrix: multiply corresponding entries and add.
\( \displaystyle \begin{bmatrix}2x & 4\end{bmatrix}\begin{bmatrix}x\\-8\end{bmatrix}=\begin{bmatrix}2x\cdot x+4(-8)\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}2x^2-32\end{bmatrix} \)
\( \displaystyle 2x^2-32=0 \)
\( \displaystyle x^2=16 \)
So x = 4 or x = −4 (both roots of \( x^2=16 \)).
Checking the statements
(A) −4 — Correct. \( 2(16)-32=0 \).
(B) 4 — Correct. \( 2(16)-32=0 \).
(C) 16 — Incorrect. \( 2(256)-32=480\ne 0 \); 16 is the value of \( x^2 \), not of x.
(D) −16 — Incorrect. \( 2(256)-32=480\ne 0 \).