Question

4x + 2y + 3z. = 35
x + 3y + 2z. = 45
2x + y + 5z. = 28

Calculat value of x,y & z
Crsmer's rule.

Answer 1

Answer on Question #49791 – Math – Matrix | Tensor Analysis

$$\begin{array}{l} 4x + 2y + 3z = 35 \\ x + 3y + 2z = 45 \\ 2x + y + 5z = 28 \\ \end{array}$$

Calculate $x, y, z$ by Cramer's Rule.

Solution:

Coefficient matrix

$$\left( \begin{array}{ccc} 4 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 5 \end{array} \right)$$

And additional column

$$\left( \begin{array}{c} 35 \\ 45 \\ 28 \end{array} \right)$$

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

$$D = \left| \begin{array}{ccc} 4 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 5 \end{array} \right|$$ $$D_x = \left| \begin{array}{ccc} 35 & 2 & 3 \\ 45 & 3 & 2 \\ 28 & 1 & 5 \end{array} \right|$$

Similarly, then $D_y$ and $D_z$ would be as follows:

$$D_y = \left| \begin{array}{ccc} 4 & 35 & 3 \\ 1 & 45 & 2 \\ 2 & 28 & 5 \end{array} \right|$$ $$D_z = \left| \begin{array}{ccc} 4 & 2 & 35 \\ 1 & 3 & 45 \\ 2 & 1 & 28 \end{array} \right|$$

Evaluating each determinant, we get:

$$\begin{array}{l} D = \left| \begin{array}{ccc} 4 & 2 & 3 \\ 1 & 3 & 2 \\ 2 & 1 & 5 \end{array} \right| = \text{expand the determinant by minors using the first column} = \\ = 4 \left| \begin{array}{cc} 3 & 2 \\ 1 & 5 \end{array} \right| - \left| \begin{array}{cc} 2 & 3 \\ 1 & 5 \end{array} \right| + 2 \left| \begin{array}{cc} 2 & 3 \\ 3 & 2 \end{array} \right| = \\ = 4 * (3 * 5 - 1 * 2) - (2 * 5 - 1 * 3) + 2 (2 * 2 - 3 * 3) \\ = 4 * (15 - 2) - (10 - 3) + 2 * (4 - 9) = 52 - 7 - 10 = 35 \end{array}$$ $$\begin{array}{l} D_x = \left| \begin{array}{ccc} 35 & 2 & 3 \\ 45 & 3 & 2 \\ 28 & 1 & 5 \end{array} \right| = \text{expand the determinant by minors using the first column} \\ = 35 * (15 - 2) - 45 * (10 - 3) + 28 * (4 - 9) = 455 - 315 - 140 = 0 \end{array}$$ $$\begin{array}{l} D_y = \left| \begin{array}{ccc} 4 & 35 & 3 \\ 1 & 45 & 2 \\ 2 & 28 & 5 \end{array} \right| = \text{expand the determinant by minors using the first column} \\ = 4 * (225 - 56) - 1 * (175 - 84) + 2 (70 - 135) = 676 - 91 - 130 = 455 \end{array}$$ $$\begin{array}{l} D_z = \left| \begin{array}{ccc} 4 & 2 & 35 \\ 1 & 3 & 45 \\ 2 & 1 & 28 \end{array} \right| = \text{expand the determinant by minors using the first column} \\ = 4 * (84 - 45) - 1 * (56 - 35) + 2 * (90 - 105) = 156 - 21 - 30 = 105 \end{array}$$

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

That is

$$x = \frac{0}{35} = 0$$ $$y = \frac{455}{35} = 13$$ $$z = \frac{105}{35} = 3$$

Answer: $\{x, y, z\} = \{0, 13, 3\}$

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