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Mathematics

If A = [3140],B=[1−223] and 3A - 5B + 2X=[4301]\begin{bmatrix}[r] 3 & 1 \ 4 & 0 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & -2 \ 2 & 3 \end{bmatrix} \text{ and 3A - 5B + 2X} = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix}; find the matrix X.

Matrices

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Answer

Given,

3A - 5B + 2X=[4301]\text{3A - 5B + 2X} = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix}

Substituting value of A and B in above equation we get,

⇒3[3140]−5[1−223]+2X=[4301]⇒[93120]−[5−101015]+2X=[4301]⇒[9−53−(−10)12−100−15]+2X=[4301]⇒[4132−15]+2X=[4301]⇒2X=[4301]−[4132−15]⇒2X=[4−43−130−21−(−15)]⇒2X=[0−10−216]⇒X=12[0−10−216]⇒X=[0−5−18].\Rightarrow 3\begin{bmatrix}[r] 3 & 1 \ 4 & 0 \end{bmatrix} - 5\begin{bmatrix}[r] 1 & -2 \ 2 & 3 \end{bmatrix} + 2X = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 & 3 \ 12 & 0 \end{bmatrix} - \begin{bmatrix}[r] 5 & -10 \ 10 & 15 \end{bmatrix} + 2X = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 - 5 & 3 - (-10) \ 12 - 10 & 0 - 15 \end{bmatrix} + 2X = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 13 \ 2 & -15 \end{bmatrix} + 2X = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] 4 & 3 \ 0 & 1 \end{bmatrix} - \begin{bmatrix}[r] 4 & 13 \ 2 & -15 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] 4 - 4 & 3 - 13 \ 0 - 2 & 1 - (-15) \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] 0 & -10 \ -2 & 16 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix}[r] 0 & -10 \ -2 & 16 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] 0 & -5 \ -1 & 8 \end{bmatrix}.

Hence, X = [0−5−18].\begin{bmatrix}[r] 0 & -5 \ -1 & 8 \end{bmatrix}.

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