Question

Let $A=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]$ Then, the system of linear equations $A^8\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right]$

• A. No solution
• B. Exactly two solutions
• C. A unique solution
• D. Infinitely many solutions

Answer 1

The correct option is A

No solution

Explanation for the correct answer:

Finding the value of $x$and $y$:

Let $A=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]$

Then, the system of linear equations

$A^8\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right]$

Given that, $A=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]$

$A^2=\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]\left[\begin{array}{cc}i& -i\\ -i& i\end{array}\right]$

{Taking i as common and mutiplying the matrix}

. $=\left[\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right]$

Taking 2 common

$=2\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]$

Now,

$$\begin{gathered} A^4=A^2·A^2 \\ =4\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right] \\ =4\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right] \\ =8\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right] \end{gathered}$$

Also,

$$\begin{gathered} A^8=A^4·A^4 \\ =64\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right] \\ =64\left[\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right] \\ =128\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right] \end{gathered}$$

Now, substituting the value of $A^8$ in $A^8\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right]$ we get,

$$\begin{gathered} \Rightarrow 128\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right] \\ \Rightarrow 128\left[\begin{array}{cc}x& -y\\ -x& +y\end{array}\right]=\left[\begin{array}{c}8\\ 64\end{array}\right] \end{gathered}$$

Converting matrix into linear equations and solving it, we get

$$\begin{gathered} \Rightarrow 128\left(x-y\right)=8,128\left(-\mathrm{x}+\mathrm{y}\right)=64 \\ \Rightarrow \mathrm{x}-\mathrm{y}=\frac{1}{16}...\left(1\right)\mathrm{x}-\mathrm{y}=-\frac{1}{2}...\left(2\right) \end{gathered}$$

From both equations, we see that there is no solution

Hence, the correct answer is option (A).