Question

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 value of α for which ${A^2} = B$ , is
A) 1
B) -1
C) 4
D) No real values

Answer 1

Hint:
First of all multiply the matrix $A$ by itself to get the matrix ${A^2}$ then substitute the value of ${A^2}$ in the given equation and also substitute the value of $B$ as the left and right hand side of the equation are equal so the matrices written on the left and right hand side is also equal .solve the equations and get the value of α.

Complete step by step solution:
We have given the equation in the above problem as:
${A^2} = B.......(1)$
The matrix A and B are also given as:
$A = \left( {\begin{array}{*{20}{c}}
  \alpha &0 \\
  1&1
\end{array}} \right)$
$B = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  5&1
\end{array}} \right)$
In the equation ${A^2}$ is given so we are going to find ${A^2}$ matrix by multiplying A by itself.
Multiplying matrix A by itself we get,
$A.A = \left( {\begin{array}{*{20}{c}}
  \alpha &0 \\
  1&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  \alpha &0 \\
  1&1
\end{array}} \right)$
$ \Rightarrow {A^2} = \left( {\begin{array}{*{20}{c}}
  {\alpha (\alpha ) + 0(1)}&{\alpha (0) + 0(1)} \\
  {1(\alpha ) + 1(1)}&{1(0) + 1(1)}
\end{array}} \right)$
$ \Rightarrow {A^2} = \left( {\begin{array}{*{20}{c}}
  {{\alpha ^2}}&0 \\
  {\alpha + 1}&1
\end{array}} \right)$
Now, substituting value of ${A^2}$and B in equation (1)

$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {{\alpha ^2}}&0 \\
  {\alpha + 1}&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  5&1
\end{array}} \right)$
Here ${\alpha ^2} = 1$ and $\alpha + 1 = 5$
Which is not possible at the same time.
So, the answer is no real values.

Note:
Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint.