MCQ
If $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$ and $A=B^2$, then $x$ equals
- A$\pm 1$
- B-1
- C1
- D2
✓
Answer
Wehave, $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}x & 0 \\ 1 & 1\end{array}\right]$
\[\therefore B^2=\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
Now, it is given that $A=B^2$
\[\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
On comparing, we get
\[\begin{array}{ll}
& x^2=1 \text { and } x+1=2 \Rightarrow x= \pm 1 \text { and } x=1 \\
\therefore & x=1
\end{array}\]
\[\therefore B^2=\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
1 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
Now, it is given that $A=B^2$
\[\Rightarrow\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]=\left[\begin{array}{cc}
x^2 & 0 \\
x+1 & 1
\end{array}\right]\]
On comparing, we get
\[\begin{array}{ll}
& x^2=1 \text { and } x+1=2 \Rightarrow x= \pm 1 \text { and } x=1 \\
\therefore & x=1
\end{array}\]
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