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Mathematics

Find the matrix X of order 2 x 2 which satisfies the equation :

[3724][0253]+2X=[1−5−46].\begin{bmatrix}[r] 3 & 7 \ 2 & 4 \end{bmatrix} \begin{bmatrix}[r] 0 & 2 \ 5 & 3 \end{bmatrix} + 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix}.

Matrices

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Answer

Given,

[3724][0253]+2X=[1−5−46],⇒[3×0+7×53×2+7×32×0+4×52×2+4×3]+2X=[1−5−46]⇒[0+356+210+204+12]+2X=[1−5−46]⇒[35272016]+2X=[1−5−46]⇒2X=[1−5−46]−[35272016]⇒2X=[1−35−5−27−4−206−16]⇒2X=[−34−32−24−10]⇒X=12[−34−32−24−10]X=[−17−16−12−5].\begin{bmatrix}[r] 3 & 7 \ 2 & 4 \end{bmatrix} \begin{bmatrix}[r] 0 & 2 \ 5 & 3 \end{bmatrix} + 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix}, \\[1em] \Rightarrow \begin{bmatrix}[r] 3 \times 0 + 7 \times 5 & 3 \times 2 + 7 \times 3 \ 2 \times 0 + 4 \times 5 & 2 \times 2 + 4 \times 3 \end{bmatrix} + 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 + 35 & 6 + 21 \ 0 + 20 & 4 + 12 \end{bmatrix} + 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 35 & 27 \ 20 & 16 \end{bmatrix} + 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] 1 & -5 \ -4 & 6 \end{bmatrix} - \begin{bmatrix}[r] 35 & 27 \ 20 & 16 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] 1 - 35 & -5 - 27 \ -4 - 20 & 6 - 16 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] -34 & -32 \ -24 & -10 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{2} \begin{bmatrix}[r] -34 & -32 \ -24 & -10 \end{bmatrix} \\[1em] X = \begin{bmatrix}[r] -17 & -16 \ -12 & -5 \end{bmatrix}.

Hence, the matrix X = [−17−16−12−5]\begin{bmatrix}[r] -17 & -16 \ -12 & -5 \end{bmatrix}.

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