Question
If \[A = \left[ \begin{array}{l}
\alpha \,\,\,\,\,\,0\\
1\,\,\,\,\,\,\,\,1
\end{array} \right]{\rm{ and }}\,\,B = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,0\\
5\,\,\,\,\,\,\,1
\end{array} \right],\] then value of $$\alpha $$ for which $${A^2} = B,$$ is
A.
1
B.
$$- 1$$
C.
4
D.
no real values
Answer :
no real values
Solution :
\[{\rm{Given\,\, that }}\,\,A = \left[ \begin{array}{l}
\alpha \,\,\,\,\,\,0\\
1\,\,\,\,\,\,\,\,1
\end{array} \right]{\rm{ and}}\,{\rm{ }}B = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,0\\
5\,\,\,\,\,\,\,1
\end{array} \right]\]
$${\text{and }}{A^2} = B$$
\[\begin{array}{l}
\Rightarrow \,\,\left[ \begin{array}{l}
\alpha \,\,\,\,\,\,0\\
1\,\,\,\,\,\,\,\,1
\end{array} \right]\left[ \begin{array}{l}
\alpha \,\,\,\,\,\,0\\
1\,\,\,\,\,\,\,\,1
\end{array} \right] = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,0\\
5\,\,\,\,\,\,\,\,1
\end{array} \right]\\
\Rightarrow \,\,\left[ \begin{array}{l}
{\alpha ^2}\,\,\,\,\,\,\,\,\,\,\,\,0\\
\alpha + 1\,\,\,\,\,\,\,\,1
\end{array} \right] = \left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,0\\
5\,\,\,\,\,\,\,\,1
\end{array} \right]
\end{array}\]
$$\eqalign{
& \Rightarrow \,\,{\alpha ^2} = 1\,\,{\text{and }}\alpha + 1 = 5 \cr
& \Rightarrow \,\,\alpha = \pm 1\,\,{\text{and }}\alpha = 4 \cr} $$
$$\because $$ There is no common value
∴ There is no real value of $$\alpha $$ for which $${A^2} = B$$