If A =[0110],\text{If A }= \begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix},If A =[0110], then A2 =
[1100]\begin{bmatrix}[r] 1 & 1 \ 0 & 0 \end{bmatrix}[1010]
[0011]\begin{bmatrix}[r] 0 & 0 \ 1 & 1 \end{bmatrix}[0101]
[0110]\begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix}[0110]
[1001]\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}[1001]
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Given,
A=[0110]⇒A2=[0110][0110]=[0×0+1×10×1+1×01×0+0×11×1+0×0]=[1001] ∴A2=[1001].\text{A} = \begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix} \\[1em] \Rightarrow \text{A}^2 = \begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix}\begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 \times 0 + 1 \times 1 & 0 \times 1 + 1 \times 0 \ 1 \times 0 + 0 \times 1 & 1 \times 1 + 0 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em]\ \therefore \text{A}^2 = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}.A=[0110]⇒A2=[0110][0110]=[0×0+1×11×0+0×10×1+1×01×1+0×0]=[1001] ∴A2=[1001].
∴ Option 4 is the correct option.
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If [x+2y3y4x2]=[0−382],\begin{bmatrix}[r] x + 2y & 3y \ 4x & 2 \end{bmatrix} = \begin{bmatrix}[r] 0 & -3 \ 8 & 2 \end{bmatrix},[x+2y4x3y2]=[08−32], then the value of x - y is
If x[23]+y[−10]=[106],x\begin{bmatrix}[r] 2 \ 3 \end{bmatrix} + y\begin{bmatrix}[r] -1 \ 0 \end{bmatrix} = \begin{bmatrix}[r] 10 \ 6 \end{bmatrix},x[23]+y[−10]=[106], then the values of x and y are
If A =[0010],\text{If A }= \begin{bmatrix}[r] 0 & 0 \ 1 & 0 \end{bmatrix},If A =[0100], then A2 =
If A =[1011],\text{If A }= \begin{bmatrix}[r] 1 & 0 \ 1 & 1 \end{bmatrix},If A =[1101], then A2 =
[2011]\begin{bmatrix}[r] 2 & 0 \ 1 & 1 \end{bmatrix}[2101]
[1012]\begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix}[1102]
[1021]\begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix}[1201]
none of these
