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If matrix M = [13−25]\begin{bmatrix}[r] 1 & 3 \ -2 & 5 \end{bmatrix}, then the matrix -2M + 2I is :

  1. [0−64−8]\begin{bmatrix}[r] 0 & -6 \ 4 & -8 \end{bmatrix}

  2. [0−468]\begin{bmatrix}[r] 0 & -4 \ 6 & 8 \end{bmatrix}

  3. [4120−4]\begin{bmatrix}[r] 4 & 12 \ 0 & -4 \end{bmatrix}

  4. [−40−86]\begin{bmatrix}[r] -4 & 0 \ -8 & 6 \end{bmatrix}

Matrices

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Answer

I = [1001]\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}

Solving, -2M + 2I

⇒−2×[13−25]+2[1001]⇒[−2−64−10]+[2002]⇒[−2+2−6+04+0−10+2]⇒[0−64−8]\Rightarrow -2 \times \begin{bmatrix}[r] 1 & 3 \ -2 & 5 \end{bmatrix} + 2\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -2 & -6 \ 4 & -10 \end{bmatrix} + \begin{bmatrix}[r] 2 & 0 \ 0 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -2 + 2 & -6 + 0 \ 4 + 0 & -10 + 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 & -6 \ 4 & -8 \end{bmatrix}

Hence, Option 1 is the correct option.

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