Question

If \[\text{A=}\left[ \begin{matrix}
   0 & 2 \\
   5 & -2 \\
\end{matrix} \right]\] , \[\text{B=}\left[ \begin{matrix}
   1 & -1 \\
   3 & 2 \\
\end{matrix} \right]\] and I is a unit matrix of order \[2\times 2\] . Find: \[{{\text{B}}^{2}}\text{A}\]

Answer 1

Hint: For the given equation we are given to find \[{{\text{B}}^{2}}\text{A}\] for the given two matrices. For this we have to do matrix multiplication. By observing the problem we can see that we have to do two matrix multiplication in the problem. For doing any type of matrix multiplication it should satisfy the order condition.

Complete step by step answer:
For solving this question we are given two matrices \[\text{A=}\left[ \begin{matrix}
   0 & 2 \\
   5 & -2 \\
\end{matrix} \right]\] and \[\text{B=}\left[ \begin{matrix}
   1 & -1 \\
   3 & 2 \\
\end{matrix} \right]\]. Now we have to find \[{{\text{B}}^{2}}\text{A}\].
To solve the problem we have to do matrix multiplication which means it should satisfy the condition that the number of columns in the first matrix must be equal to the number of rows in the second matrix.
Let us check the above condition for given matrix, to solve this problem we have to do operations like \[{{B}^{2}}\] and \[{{B}^{2}}A\], for \[B\times B\] we have orders \[2\times 2\] and \[2\times 2\] respectively so it satisfies the condition.
Now for finding \[{{\text{B}}^{2}}\text{A}\] we have to find \[{{\text{B}}^{2}}\] and then we have to multiply with A.
First of all let us find \[{{\text{B}}^{2}}\]. For that we have to multiply B matrix with B matrix i.e. matrix multiplication.
As we know multiplication of matrix for the any matrix \[A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\] and \[B=\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]\] will be.
\[AB=\left[ \begin{matrix}
   ae+bg & af+bh \\
   ce+dg & cf+dh \\
\end{matrix} \right]\]
Let us apply the above concept to the A and B matrices, we get
\[\begin{align}
  & \Rightarrow B{{}^{2}}=\left[ \begin{matrix}
   1 & -1 \\
   3 & 2 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   1 & -1 \\
   3 & 2 \\
\end{matrix} \right] \\
 & \Rightarrow \text{ =}\left[ \begin{matrix}
   1\times 1-1\times 3 & 1\times -1+-1\times 2 \\
   3\times 1+2\times 3 & 3\times 1+2\times 2 \\
\end{matrix} \right] \\
 & \Rightarrow \text{ =}\left[ \begin{matrix}
   -2 & -4 \\
   9 & 7 \\
\end{matrix} \right] \\
\end{align}\]
Now we have to do the operation \[{{\text{B}}^{2}}\times \text{A}\]
By applying the above concept we get:
\[\begin{align}
  & \Rightarrow {{B}^{2}}\times A=\left[ \begin{matrix}
   -2 & -4 \\
   9 & 7 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   0 & 2 \\
   5 & -2 \\
\end{matrix} \right] \\
 & \Rightarrow \text{ =}\left[ \begin{matrix}
   -2\times 0-4\times 5 & -2\times 2+4\times 2 \\
   9\times 0+7\times 5 & 9\times 2-7\times 2 \\
\end{matrix} \right] \\
 & \Rightarrow \text{ =}\left[ \begin{matrix}
   20 & 4 \\
   35 & 14 \\
\end{matrix} \right] \\
\end{align}\]
Hence \[{{\text{B}}^{2}}\text{A}\]is \[\text{=}\left[ \begin{matrix}
   20 & 4 \\
   35 & 14 \\
\end{matrix} \right]\].

Note: For any square matrix, matrix multiplication will be easy because order condition will always be satisfied whereas for a matrix with order \[m\times n\] we should check the order condition for every multiplication. Students should avoid calculation mistakes while solving these problems.