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Question

Solve the following system of equations by matrix method.
$$x-y+z=4, 2x+y-3z=0$$ and $$x+y+z=2$$.


Solution

We first write the given system of equations in matrix form$$(AX=B)$$ and then solve it.
$$\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$$
We reduce the matrix $$A$$ in identity matrix by performing row operations to determine the variables
$$R_2\rightarrow R_2-2R_1$$ and $$R_3\rightarrow R_3-R_1$$ we get

$$\begin{bmatrix} 1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 4 \\ -8\\ -2 \end{bmatrix}$$
$$R_1\rightarrow R_1+\dfrac{1}{3}R_2$$ and $$R_3\rightarrow R_3-\dfrac{2}{3}R_2,$$ we get

$$\begin{bmatrix} 1 & 0 & -\frac{2}{3} \\ 0 & 3 & -5 \\ 0 & 0 & \frac{10}{3}\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} \frac{4}{3} \\ -8 \\ \frac{10}{3} \end{bmatrix}$$
$$R_2\rightarrow \dfrac{R_2}{3}$$ we get

$$\begin{bmatrix} 1 & 0 & -\frac{2}{3} \\ 0 & 1 & -\frac{5}{3} \\ 0 & 0 & \frac{10}{3}\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} \frac{4}{3} \\ -\frac{8}{3} \\ \frac{10}{3} \end{bmatrix}$$
$$R_1\rightarrow R_1 + \dfrac{1}{5} R_3$$ and $$R_2\rightarrow R_2+\dfrac{1}{2}R_3$$ we get


$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & \frac{10}{3}\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ \frac{10}{3} \end{bmatrix}$$
$$R_3\rightarrow R_3\times \dfrac{3}{10},$$ we get

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ 1  \end{bmatrix}$$

$$\therefore$$ $$x=2, y=-1$$ and $$z=1$$

Maths

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