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Mathematics Question on Matrices

If A=[10 11]A = \begin{bmatrix}1&0\\\ 1&1\end{bmatrix} and A8=aA+bI,A^8 = aA +bI, then (a,b)=(a , b) =

A

(8,7)(8, 7)

B

(−7,8)(-7, 8)

C

(8,−7)(8, -7)

D

(−8,−7)(-8, -7)

Answer

(8,−7)(8, -7)

Explanation

Solution

We have, A=[10 11]A = \begin{bmatrix}1&0\\\ 1&1\end{bmatrix}
A2=A.A=[10 11][10 11]=[10\21]A^2 = A . A = \begin{bmatrix}1&0\\\ 1&1\end{bmatrix} \begin{bmatrix}1&0\\\ 1&1\end{bmatrix} = \begin{bmatrix}1&0\\\2&1\end{bmatrix}
A3=A2.A=[10\21][10 11]=[10 31]A^3 = A^2 . A = \begin{bmatrix}1&0\\\2&1\end{bmatrix} \begin{bmatrix}1&0\\\ 1&1\end{bmatrix} = \begin{bmatrix}1&0\\\ 3&1\end{bmatrix}
Similarly , A8=[10\81] A^8 = \begin{bmatrix}1&0\\\8 &1\end{bmatrix}
Now, A8=aA+bI A^8 = aA + bI
⇒   [10 81]=a[10 11]+b[10 01]\Rightarrow \:\:\: \begin{bmatrix}1&0\\\ 8 &1\end{bmatrix} = a \begin{bmatrix}1&0\\\ 1&1\end{bmatrix} + b \begin{bmatrix}1 &0\\\ 0 & 1\end{bmatrix}
⇒   [10 81]=[a+b0 aa+b]\Rightarrow \:\:\: \begin{bmatrix}1&0\\\ 8&1\end{bmatrix} = \begin{bmatrix} a+ b &0\\\ a & a+b \end{bmatrix}
∴    a=8\therefore \:\:\:\: a = 8 and a+b=1  ⇒ b=1−8=−7a + b = 1 \:\: \Rightarrow \: b = 1 - 8 = - 7