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3 and 4 .Determinants and Matrices
normal
If $A = \left[ {\begin{array}{*{20}{c}}
\alpha &0\\
1&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}
1&0\\
5&1
\end{array}} \right]$ , then the value of $\alpha $ for which $A^2 = B$ is
A
$1$
B
$-1$
C
$4$
D
No real values
Solution
$A=\left[\begin{array}{ll}{\alpha} & {0} \\ {1} & {1}\end{array}\right]$
$A^{2}=\left[\begin{array}{ll}{\alpha} & {0} \\ {1} & {1}\end{array}\right]\left[\begin{array}{ll}{\alpha} & {0} \\ {1} & {1}\end{array}\right]$
$=\left[\begin{array}{cc}{\alpha^{2}} & {0} \\ {\alpha+1} & {1}\end{array}\right]$
$\alpha^{2}=1 \Rightarrow \alpha=\pm 1$
$\alpha+1=5 \Rightarrow \alpha=4$
$\left\langle {no\,\,common\,\,value\,\,of\,\alpha } \right\rangle $
so no real $\alpha$
Standard 12
Mathematics
