Question

If we are given a matrix as \[A=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right]\] and \[{{\left( a{{I}_{2}}+bA \right)}^{2}}=A\] then the values of \[a,b\] are:
(a) \[a=b=\sqrt{2}\]
(b) \[a=b=\dfrac{1}{\sqrt{2}}\]
(c) \[a=b=\sqrt{3}\]
(d) \[a=b=\dfrac{1}{\sqrt{3}}\]

Answer 1

Hint: We solve this problem by substituting the given matrix in the given equation.
We use the identity matrix that is
\[{{I}_{2}}=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
We have the formula of multiplication of scalar with matrix that is
\[x\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]=\left[ \begin{matrix}
   xp & xq \\
   xr & xs \\
\end{matrix} \right]\]
Then we use the square of the matrix that is multiplying the matrix with itself. The multiplication of the matrix with itself is done by using the dot product of rows to columns.
The multiplication of matrix \[B=\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]\] with itself can be done as
\[\Rightarrow {{B}^{2}}=\left[ \begin{matrix}
   {{p}^{2}}+qr & pq+qs \\
   rp+rs & rq+{{s}^{2}} \\
\end{matrix} \right]\]
Then we use the condition that if two matrices are equal and only of every corresponding element in two matrices are equal.

Complete step-by-step solution
We are given that the matrix as
\[A=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right]\]
We also given with the equation that is
\[{{\left( a{{I}_{2}}+bA \right)}^{2}}=A\]
We know that the identity matrix that is
\[{{I}_{2}}=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
By substituting the required matrices in the given equation we get
\[\Rightarrow {{\left( a\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]+b\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right] \right)}^{2}}=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right]\]
We know that the formula of multiplication of scalar with matrix that is
\[x\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]=\left[ \begin{matrix}
   xp & xq \\
   xr & xs \\
\end{matrix} \right]\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow {{\left( \left[ \begin{matrix}
   a & 0 \\
   0 & a \\
\end{matrix} \right]+\left[ \begin{matrix}
   0 & b \\
   -b & 0 \\
\end{matrix} \right] \right)}^{2}}=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right] \\
 & \Rightarrow {{\left( \left[ \begin{matrix}
   a & b \\
   -b & a \\
\end{matrix} \right] \right)}^{2}}=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right] \\
\end{align}\]
We know that the square of a matrix is obtained by multiplying the matrix with itself so we get
\[\Rightarrow \left[ \begin{matrix}
   a & b \\
   -b & a \\
\end{matrix} \right]\times \left[ \begin{matrix}
   a & b \\
   -b & a \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right]\]
We know that the multiplication of matrix \[B=\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]\] with itself can be done as
\[\Rightarrow {{B}^{2}}=\left[ \begin{matrix}
   {{p}^{2}}+qr & pq+qs \\
   rp+rs & rq+{{s}^{2}} \\
\end{matrix} \right]\]
By using this formula of multiplication of matrices to above equation we get
\[\Rightarrow \left[ \begin{matrix}
   {{a}^{2}}-{{b}^{2}} & 2ab \\
   -2ab & {{a}^{2}}-{{b}^{2}} \\
\end{matrix} \right]=\left[ \begin{matrix}
   0 & 1 \\
   -1 & 0 \\
\end{matrix} \right]\]
We know that the condition that if two matrices are equal and only every corresponding element in the two matrices are equal.
By using this theorem to first element in the above equation we get
\[\begin{align}
  & \Rightarrow {{a}^{2}}-{{b}^{2}}=0 \\
 & \Rightarrow a=\pm b \\
\end{align}\]
Now, by taking the second element we get
\[\Rightarrow 2ab=1\]
Now, by substituting the value of \['a'\] in above equation we get
\[\begin{align}
  & \Rightarrow 2b\left( \pm b \right)=1 \\
 & \Rightarrow {{b}^{2}}=\mp \dfrac{1}{2} \\
\end{align}\]
Here we can see that the negative sign will not hold because we know that the square of a number cannot be negative. So, by taking the positive sign we get
\[\begin{align}
  & \Rightarrow {{b}^{2}}=\dfrac{1}{2} \\
 & \Rightarrow b=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}\]
Therefore we can conclude that the values of \[a,b\] are
\[\Rightarrow a=b=\pm \dfrac{1}{\sqrt{2}}\]
Here we can get two possibilities one by taking the positive sign and other by negative sign so, we get the values of \[a,b\] as
\[\begin{align}
  & \therefore a=b=\dfrac{1}{\sqrt{2}} \\
 & \therefore a=b=-\dfrac{1}{\sqrt{2}} \\
\end{align}\]
So, option (b) is the correct answer.

Note: Students may make mistakes in the multiplication of a scalar with the matrix.
We have the formula of multiplication of scalar with matrix as
\[x\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]=\left[ \begin{matrix}
   xp & xq \\
   xr & xs \\
\end{matrix} \right]\]
But students may do mistake and multiply the scalar to only one column in the matrix which is a property of determinant, that is
\[x\left[ \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right]=\left[ \begin{matrix}
   xp & q \\
   xr & s \\
\end{matrix} \right]\]
This gives the wrong answer because the multiplication of a scalar to matrix is done to all elements.
Also we have the value of \['b'\] as
\[\Rightarrow b=\pm \dfrac{1}{\sqrt{2}}\]
Here, students may miss the \[\pm \] sign which gives one answer less.
For the multiple choice questions this is very important to take the \[\pm \] sign because we may get more than one answer if we consider the\[\pm \] sign.