Let A = begin{bmatrix} i & -i -i & i end{bma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} i & -i \ -i & i \end{bmatrix} \begin{

Vedclass · Accepted Answer

No solution

Vedclass · Answer

(C) Given $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} i & -i \ -i & i \end{bmatrix} \begin{bmatrix} i & -i \ -i & i \end{bmatrix} = \begin{bmatrix} i^{2} + (-i)(-i) & -i^{2} - i^{2} \ -i^{2} - i^{2} & (-i)(-i) + i^{2} \end{bmatrix} = \begin{bmatrix} -1 - 1 & 1 + 1 \ 1 + 1 & -1 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix} = 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}$.Next,calculate $A^{4} = (A^{2})^{2}$:$A^{4} = \left( 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} ight)^{2} = 4 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} = 4 \begin{bmatrix} 1+1 & -1-1 \ -1-1 & 1+1 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$.Then,calculate $A^{8} = (A^{4})^{2}$:$A^{8} = \left( 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} ight)^{2} = 64 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} = 64 \begin{bmatrix} 1+1 & -1-1 \ -1-1 & 1+1 \end{bmatrix} = 64 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$.The system is $128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 64 \end{bmatrix}$.This implies $128(x - y) = 8 \Rightarrow x - y = \frac{8}{128} = \frac{1}{16}$ and $128(-x + y) = 64 \Rightarrow -x + y = \frac{64}{128} = \frac{1}{2}$.From the first equation,$x - y = \frac{1}{16}$,and from the second,$x - y = -\frac{1}{2}$.Since $\frac{1}{16} 
eq -\frac{1}{2}$,the system has no solution.

Let $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$,where $i = \sqrt{-1}$. Then,the system of linear equations $A^{8} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$ has :

Similar Questions

Consider the system of equations: $x + ay = 0$,$y + az = 0$,and $z + ax = 0$. The set of all real values of $a$ for which the system has a unique solution is:

If the system of equations $x+y+2z=3$, $x+2y+3z=4$ and $x+y+cz=5$ is inconsistent, then:

If the system of linear equations $2x + 3y - z = -2$; $x + y + z = 4$; $x - y + |\lambda|z = 4\lambda - 4$ (where $\lambda \in R$) has no solution,then:

The following system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,$x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$ for $\lambda$ equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module