Question

Solve the system of simultaneous equations below using any of the matrix method
2x + 3y - z = 6
3x + y + 5z =8
x + 2y + z =10

Answer 1

Answer on Question #51643 - Math - Matrix | Tensor Analysis

Solve the system of simultaneous equations below using any of the matrix method

$$\begin{array}{l} 2x + 3y - z = 6 \\ 3x + y + 5z = 8 \\ x + 2y + z = 10 \\ \end{array}$$

Solution

Cramer's rule will be applied to calculate $x, y, z$.

Coefficient matrix

$$\left( \begin{array}{ccc} 2 & 3 & -1 \\ 3 & 1 & 5 \\ 1 & 2 & 1 \end{array} \right)$$

and answer column

$$\left( \begin{array}{c} 6 \\ 8 \\ 10 \end{array} \right)$$

We have the left-hand side of the system with the variables (the "coefficient matrix") and the right-hand side with the answer values.

Let $D$ be the determinant of the coefficient matrix of the above system, and let $D_x$ be the determinant formed by replacing the $x$-column values with the answer-column values

$$D = \left| \begin{array}{ccc} 2 & 3 & -1 \\ 3 & 1 & 5 \\ 1 & 2 & 1 \end{array} \right|$$ $$D_x = \left| \begin{array}{ccc} 6 & 3 & -1 \\ 8 & 1 & 5 \\ 10 & 2 & 1 \end{array} \right|$$

Similarly, $D_y$ and $D_z$ would then be the following:

$$D_y = \left| \begin{array}{ccc} 2 & 6 & -1 \\ 3 & 8 & 5 \\ 1 & 10 & 1 \end{array} \right|$$ $$D_z = \left| \begin{array}{ccc} 2 & 3 & 6 \\ 3 & 1 & 8 \\ 1 & 2 & 10 \end{array} \right|$$

Evaluating each determinant, we get:

$$\begin{array}{l} D = \left| \begin{array}{ccc} 2 & 3 & -1 \\ 3 & 1 & 5 \\ 1 & 2 & 1 \end{array} \right| = | \text{expand along the first column} | = 2 \left| \begin{array}{cc} 1 & 5 \\ 2 & 1 \end{array} \right| - 3 \left| \begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array} \right| + \left| \begin{array}{cc} 3 & -1 \\ 1 & 5 \end{array} \right| \\ = 2(1 - 1) - 2(1 - 5) - 3(3 - 1) + (3 - 5 - 1 - (-1)) \\ = 2(1 - 10) - 3(3 + 2) + (15 + 1) = -18 - 15 + 16 = -17 \\ \end{array}$$ $$\begin{array}{l} D_{x} = \left| \begin{array}{ccc} 6 & 3 & -1 \\ 8 & 1 & 5 \\ 10 & 2 & 1 \end{array} \right| = | \text{expand along the first column} | \\ = 6 \left| \begin{array}{cc} 1 & 5 \\ 2 & 1 \end{array} \right| - 8 \left| \begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array} \right| + 10 \left| \begin{array}{cc} 3 & -1 \\ 1 & 5 \end{array} \right| = 6(1 - 10) - 8(3 + 2) + 10(15 + 1) \\ = -54 - 40 + 160 = 66 \end{array}$$ $$\begin{array}{l} D_{y} = \left| \begin{array}{ccc} 2 & 6 & -1 \\ 3 & 8 & 5 \\ 1 & 10 & 1 \end{array} \right| = | \text{expand along the first column} | \\ = 2 \left| \begin{array}{cc} 8 & 5 \\ 10 & 1 \end{array} \right| - 3 \left| \begin{array}{cc} 6 & -1 \\ 10 & 1 \end{array} \right| + \left| \begin{array}{cc} 6 & -1 \\ 8 & 5 \end{array} \right| = 2(8 - 50) - 3(6 + 10) + (30 + 8) \\ = -84 - 48 + 38 = -94 \end{array}$$ $$\begin{array}{l} D_{z} = \left| \begin{array}{ccc} 2 & 3 & 6 \\ 3 & 1 & 8 \\ 1 & 2 & 10 \end{array} \right| = | \text{expand along the first column} | = 2 \left| \begin{array}{cc} 1 & 8 \\ 2 & 10 \end{array} \right| - 3 \left| \begin{array}{cc} 3 & 6 \\ 2 & 10 \end{array} \right| + \left| \begin{array}{cc} 3 & 6 \\ 1 & 8 \end{array} \right| \\ = 2(10 - 16) - 3(30 - 12) + (24 - 6) = -12 - 54 + 18 = -48 \end{array}$$

Cramer's Rule says that $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D},$

that is,

$$x = \frac{66}{-17} = -3 \frac{15}{17}$$ $$y = \frac{-94}{-17} = 5 \frac{9}{17}$$ $$z = \frac{-48}{-17} = 2 \frac{14}{17}$$

Answer: $\{x, y, z\} = \left\{-3\frac{15}{17}, 5\frac{9}{17}, 2\frac{14}{17}\right\}$

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