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Question: If \(A=\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right]\) and \({{A}^{2}}...

If A=[10 −17 ]A=\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right] and A2=8A+KI2{{A}^{2}}=8A+K{{I}_{2}} , then KK is equal to:
A)−1A)-1
B)1B)1
C)−7C)-7
D)7D)7

Explanation

Solution

Hint : To solve this question we need to have the knowledge of matrix and its operation. Firstly we will find the value for A2{{A}^{2}}. The matrix IIdenotes the identity matrix of four elements. We will do the operation of the matrix as per given in the question. Then we will equate the elements in the Right Hand Side of the matrix to that in the Left Hand Side of the matrix.

Complete step-by-step solution:
The question ask us to find value of KK, if the condition given is A2=8A+KI2{{A}^{2}}=8A+K{{I}_{2}} and the matrix A is A=[10 −17 ]A=\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right]. The first step is to calculate A2{{A}^{2}} . On multiplying the two matrices we get:
⇒A2=[10 −17 ][10 −17 ]\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right]
⇒A2=[1×1+0×1−1×0+7×0 −1×1+7×−10×−1+7×7 ]\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 1\times 1+0\times 1 & -1\times 0+7\times 0 \\\ -1\times 1+7\times -1 & 0\times -1+7\times 7 \\\ \end{matrix} \right]
On solving the matrix we get:
⇒A2=[10 −849 ]\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\\ -8 & 49 \\\ \end{matrix} \right]
Now we will solve for the L.H.S of the expression A2=8A+KI2{{A}^{2}}=8A+K{{I}_{2}}. On solving we get:
⇒8A+KI2=8[10 −17 ]+K[10 01 ]\Rightarrow 8A+K{{I}_{2}}=8\left[ \begin{matrix} 1 & 0 \\\ -1 & 7 \\\ \end{matrix} \right]+K\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]
⇒8A+KI2=[80 −856 ]+[K0 0K ]\Rightarrow 8A+K{{I}_{2}}=\left[ \begin{matrix} 8 & 0 \\\ -8 & 56 \\\ \end{matrix} \right]+\left[ \begin{matrix} K & 0 \\\ 0 & K \\\ \end{matrix} \right]
Since the above two matrix of same length are being added so each element will be added giving the matrix as:
⇒8A+KI2=[8+K0 −856+K ]\Rightarrow 8A+K{{I}_{2}}=\left[ \begin{matrix} 8+K & 0 \\\ -8 & 56+K \\\ \end{matrix} \right]
Now the R.H.S. of the expression is [8+K0 −856+K ]\left[ \begin{matrix} 8+K & 0 \\\ -8 & 56+K \\\ \end{matrix} \right]. We will equate to [10 −849 ]\left[ \begin{matrix} 1 & 0 \\\ -8 & 49 \\\ \end{matrix} \right].On equating we will get:
⇒[10 −849 ]=[8+K0 −856+K ]\Rightarrow \left[ \begin{matrix} 1 & 0 \\\ -8 & 49 \\\ \end{matrix} \right]=\left[ \begin{matrix} 8+K & 0 \\\ -8 & 56+K \\\ \end{matrix} \right]
On equating we will get:
⇒8+K=1\Rightarrow 8+K=1
⇒K=1−8\Rightarrow K=1-8
⇒K=−7\Rightarrow K=-7
∴\therefore The value for KK is equal to C)−7C)-7.

Note: Do remember that all the matrices cannot be multiplied. For two matrices to be multiplied when the number of rows of the second matrix is the same as the number of columns of the first matrix. The above answer can be checked by substituting the value of KKin the matrix [8+K0 −856+K ]\left[ \begin{matrix} 8+K & 0 \\\ -8 & 56+K \\\ \end{matrix} \right]. ON putting K=−7K=-7in the matrix, we get
⇒[8−70 −856−7 ]\Rightarrow \left[ \begin{matrix} 8-7 & 0 \\\ -8 & 56-7 \\\ \end{matrix} \right]
⇒[10 −849 ]\Rightarrow \left[ \begin{matrix} 1 & 0 \\\ -8 & 49 \\\ \end{matrix} \right]
Since the above matrix is same as A2{{A}^{2}}, so the answer we got is correct.