Question

If matrix \[A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\] and \[I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\], find K so that \[{{A}^{2}}=KA-2I\].

Answer 1

Hint: Here in this question, we need to find out the value of K in the \[{{A}^{2}}=KA-2I\]. Before solving this, we need to look at the definition of matrix. After that, we will consider that given data and given expression, firstly in the expression, we choose \[{{A}^{2}}\] then evaluate it, and again we evaluate the right-hand side, finally, we equate both LHS and RHS and then get the value of K.

Complete step by step answer:
Matrix: A matrix is a rectangular arrangement of numbers (real or complex) which may be represented as
\[\left( \begin{matrix}
   {{a}_{11}} & \ldots & {{a}_{1n}} \\
   \vdots & \ddots & \vdots \\
   {{a}_{m1}} & \cdots & {{a}_{mn}} \\
\end{matrix} \right)\]
Matrix is enclosed by \[\left[ {} \right]\]or \[\left( {} \right)\].
 Compact from the above matrix is represented by \[{{\left[ {{a}_{ij}} \right]}_{m\times n}}\]or \[A=\left[ {{a}_{ij}} \right]\].

Let us solve our question,
Given data is \[A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\] and \[I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
Given expression is \[{{A}^{2}}=KA-2I\]
 Let us consider the given expression \[{{A}^{2}}=KA-2I\],
Firstly, we solve the left-hand side, we have \[{{A}^{2}}\]
\[{{A}^{2}}\]can be written as \[{{A}^{2}}=A\times A\].
We are going to substitute the \[A=\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\] in \[{{A}^{2}}=A\times A\], then we get
\[\Rightarrow \left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]\]
On solving,
\[\Rightarrow \left[ \begin{matrix}
   9-8 & -6+4 \\
   12-8 & -8+4 \\
\end{matrix} \right]\]
Above matrix is obtained by \[2\times 2\] matrix multiplication,
If we want to do for the first element in the first row,
We simply, calculate first row of $1^{st}$ matrix with first column of $2^{nd}$ matrix
\[\begin{align}
  & 3=\left( 3\times 3 \right)+\left( -2\times 4 \right) \\
 & \Rightarrow 9-8 \\
\end{align}\]
Similarly, we can do rest of the elements in the same manner,
Therefore,\[{{A}^{2}}=\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]\], which is the Left-hand side.
Consider whole expression, \[{{A}^{2}}=KA-2I\]
Substituting the respective matrices in the above
\[\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=K\left[ \begin{matrix}
   3 & -2 \\
   4 & -2 \\
\end{matrix} \right]-2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
On right-hand side, we are going to multiply the individual terms, then we get
\[\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3K & -2K \\
   4K & -2K \\
\end{matrix} \right]-\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right]\]
Again, we are combining the matrix on right-hand side,
 \[\left[ \begin{matrix}
   1 & -2 \\
   4 & -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   3K-2 & -2K-0 \\
   4K-0 & -2K-0 \\
\end{matrix} \right]\]
Now, we are going to find the value of K, but considering the corresponding elements on both sides,
Firstly, we are considering the first row first element of LHS matrix and equalising with first row first element of RHS matrix
\[\begin{align}
  & 3K-2=1 \\
 & \Rightarrow 3K=3 \\
 & \Rightarrow K=1 \\
\end{align}\]
Firstly, we are considering the second row first element of LHS matrix and equalising with second row first element of RHS matrix
\[\begin{align}
  & -2K=-2 \\
 & \Rightarrow K=1 \\
\end{align}\]
Firstly, we are considering the second row first element of LHS matrix and equalising with second row first element of RHS matrix

\[\begin{align}
  & 4K=4 \\
 & \Rightarrow K=1 \\
\end{align}\]
Firstly, we are considering the second row second element of LHS matrix and equalising with second row second element of RHS matrix
\[\begin{align}
  & -2K-2=-4 \\
 & \Rightarrow -2K=-2 \\
 & \Rightarrow K=1 \\
\end{align}\]
\[\therefore K=1\]

Therefore, we can say that the value of K is 1.

Note: It is important to note that when we consider two matrices to be equal then in order to hold the equality every corresponding element in both the matrices should be equal. For matrix multiplication, the number of columns present in the first matrix should be equal to the number of rows present in the second matrix.