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Mathematics

If matrix M = [3−24−2]\begin{bmatrix}[r] 3 & -2 \ 4 & -2 \end{bmatrix}, find M2 + 3I.

Matrices

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Answer

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

Solving, M2 + 3I :

⇒[3−24−2][3−24−2]+3[1001]⇒[3×3+(−2)×43×−2+(−2)×(−2)4×3+(−2)×44×−2+(−2)×(−2)]+[3003]⇒[9−8−6+412−8−8+4]+[3003]⇒[1−24−4]+[3003]⇒[1+3−2+04+0−4+3]⇒[4−24−1].\Rightarrow \begin{bmatrix}[r] 3 & -2 \ 4 & -2 \end{bmatrix}\begin{bmatrix}[r] 3 & -2 \ 4 & -2 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3 \times 3 + (-2) \times 4 & 3 \times -2 + (-2) \times (-2) \ 4 \times 3 + (-2) \times 4 & 4 \times -2 + (-2) \times (-2) \end{bmatrix} + \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 - 8 & -6 + 4 \ 12 - 8 & -8 + 4 \end{bmatrix} + \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 1 & -2 \ 4 & -4 \end{bmatrix} + \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 1 + 3 & -2 + 0 \ 4 + 0 & -4 + 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & -2 \ 4 & -1 \end{bmatrix}.

Hence, M2 + 3I = [4−24−1].\begin{bmatrix}[r] 4 & -2 \ 4 & -1 \end{bmatrix}.

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