Question

If matrix $A=\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]$ and ${{A}^{2}}=KA$, then write the value of K.

Answer 1

We are given a matrix as $A=\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]$. Find ${{A}^{2}}$ by multiply matrix $A\times A$. Then solve the matrix by expanding and applying algebraic identities. Later, try to get the answer in a form of KA where K is a constant and A is the given matrix. Hence, we get the value of K.

Complete step-by-step solution:
Since, we have a matrix given as: $A=\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]$. The matrix is of order $2\times 2$.
So, for a matrix of order $2\times 2$, say$A=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$,
We have:

& {{A}^{2}}=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} a\times a+b\times c & a\times b+b\times d \\\ c\times a+d\times c & c\times b+d\times d \\\ \end{matrix} \right] \end{aligned}$$ So, for the given matrix $A=\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]$, we can write: $$\begin{aligned} & {{A}^{2}}=\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} 1\times 1+(-1)\times (-1) & 1\times (-1)+(-1)\times 1 \\\ (-1)\times 1+1\times (-1) & (-1)\times (-1)+1\times 1 \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} 1+1 & -1-1 \\\ -1-1 & 1+1 \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} 2 & -2 \\\ -2 & 2 \\\ \end{matrix} \right]......(1) \end{aligned}$$ Now, take 2 as common from the equation (1), we get: $${{A}^{2}}=2\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]......(2)$$ Now, compare equation (2), with ${{A}^{2}}=KA$ We get: $\begin{aligned} & 2\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]=KA \\\ & 2\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right]=K\left[ \begin{matrix} 1 & -1 \\\ -1 & 1 \\\ \end{matrix} \right] \\\ & 2=K \end{aligned}$ **Hence, the value of K is 2.** **Note:** While multiplying two matrices, be careful while choosing the elements and put them in correct place according to the give formula, i.e. $$\begin{aligned} & {{A}^{2}}=\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} a\times a+b\times c & a\times b+b\times d \\\ c\times a+d\times c & c\times b+d\times d \\\ \end{matrix} \right] \end{aligned}$$ There is a chance of getting wrong answer due to error in multiplication of matrices. Also, while applying any algebraic identity to the given matrix, it gets applied to all the elements of the matrix. So, when we took 2 common from the matrix, check whether each term is a multiple of 2 or not. Since, matrices seem like determinant, do not consider it as a determinant. So, we cannot solve the matrix as we solve determinant, i.e. expanding the whole values inside the bracket and then adding them to get a finite value. The final value of a matrix is itself a matrix.