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Mathematics

If A = [1011],B=[0110] and C=[1100]\begin{bmatrix}[r] 1 & 0 \ 1 & 1 \end{bmatrix}, B = \begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix}\text{ and C} = \begin{bmatrix}[r] 1 & 1 \ 0 & 0 \end{bmatrix}, the matrix A2 + 2B - 3C is :

  1. [−2−141]\begin{bmatrix}[r] -2 & -1 \ 4 & 1 \end{bmatrix}

  2. [2−141]\begin{bmatrix}[r] 2 & -1 \ 4 & 1 \end{bmatrix}

  3. [2141]\begin{bmatrix}[r] 2 & 1 \ 4 & 1 \end{bmatrix}

  4. [21−4−1]\begin{bmatrix}[r] 2 & 1 \ -4 & -1 \end{bmatrix}

Matrices

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Answer

Substituting values of A, B and C in A2 + 2B - 3C, we get :

⇒A2+2B−3C=[1011][1011]+2[0110]−3[1100]=[1×1+0×11×0+0×11×1+1×11×0+1×1]+[0220]−[3300]=[1+00+01+10+1]+[0220]−[3300]=[1021]+[0220]−[3300]=[1+0−30+2−32+2−01+0−0]=[−2−141]\Rightarrow A^2 + 2B - 3C = \begin{bmatrix}[r] 1 & 0 \ 1 & 1 \end{bmatrix}\begin{bmatrix}[r] 1 & 0 \ 1 & 1 \end{bmatrix} + 2\begin{bmatrix}[r] 0 & 1 \ 1 & 0 \end{bmatrix} - 3\begin{bmatrix}[r] 1 & 1 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 1 \ 1 \times 1 + 1 \times 1 & 1 \times 0 + 1 \times 1 \end{bmatrix} + \begin{bmatrix}[r] 0 & 2 \ 2 & 0 \end{bmatrix} - \begin{bmatrix}[r] 3 & 3 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 0 & 0 + 0 \ 1 + 1 & 0 + 1 \end{bmatrix} + \begin{bmatrix}[r] 0 & 2 \ 2 & 0 \end{bmatrix} - \begin{bmatrix}[r] 3 & 3 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix} + \begin{bmatrix}[r] 0 & 2 \ 2 & 0 \end{bmatrix} - \begin{bmatrix}[r] 3 & 3 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 0 - 3 & 0 + 2 - 3 \ 2 + 2 - 0 & 1 + 0 - 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 & -1 \ 4 & 1 \end{bmatrix}

Hence, Option 1 is the correct option.

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