If \( P=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix},\ Q=\begin{bmatrix}x & 0\\1 & 1\end{bmatrix} \) and \( P=Q^2 \), then \( x \) equals:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
Since \( P=Q^2 \), the entries of \( Q^2 \) must match those of P entry by entry.
\( \displaystyle Q^2=\begin{bmatrix}x & 0\\1 & 1\end{bmatrix}\begin{bmatrix}x & 0\\1 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}x^2 & 0\\x+1 & 1\end{bmatrix} \)
\( \displaystyle \begin{bmatrix}x^2 & 0\\x+1 & 1\end{bmatrix}=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix} \)
\( \displaystyle x^2=1\ \Rightarrow\ x=\pm 1 \)
\( \displaystyle x+1=2\ \Rightarrow\ x=1 \)
Both conditions must hold, so x = 1.
Why the other options are incorrect
Option (2) \( \pm 1 \) — Incorrect. \( x^2=1 \) alone allows ±1, but the (2, 1) entry also needs x + 1 = 2, which rules out x = −1.
Option (3) \( -1 \) — Incorrect. With x = −1 the (2, 1) entry of \( Q^2 \) is 0, not 2.
Option (4) 2 — Incorrect. With x = 2 the (1, 1) entry of \( Q^2 \) is 4, not 1.