Q. 57

Question

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equation

$\{\begin{matrix} x + y + z = 9 \\ 3 x + 2 y - z = 8 \\ 3 x + y + 2 z = 1 \end{matrix}$

Step-by-Step Solution

Verified

Answer

The solution of the matrix is $x = - \frac{34}{7}, y = \frac{85}{7}, and z = \frac{12}{7}$

1Step 1. Given information

A matrix is given as,

$\{\begin{array}{r} x + y + z = 9 \\ 3 x + 2 y - z = 8 \\ 3 x + y + 2 z = 1 \end{array}$

We need to find the solution of the matrix using inverse of the matrix.

2Step 2. Inverse of the matrix.

The inverse of the matrix can be found by :

$[A ∣ I_{3}] = [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 1 & 0 \\ 3 & 1 & 2 & 0 & 0 & 1 \end{matrix}] R_{2} = \frac{r_{2}}{3} and R_{3} = \frac{r_{3}}{3} [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 3 & 2 & - 1 & 0 & 1 & 0 \\ 3 & 1 & 2 & 0 & 0 & 1 \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ \frac{3}{3} & \frac{2}{3} & \frac{- 1}{3} & \frac{0}{3} & \frac{1}{3} & \frac{0}{3} \\ \frac{3}{3} & \frac{1}{3} & \frac{2}{3} & \frac{0}{3} & \frac{0}{3} & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & \frac{2}{3} & - \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 1 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & \frac{1}{3} \end{matrix}] R_{2} = r_{2} - r_{1} [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & \frac{2}{3} & - \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 1 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 - 1 & \frac{2}{3} - 1 & - \frac{1}{3} - 1 & 0 - 1 & \frac{1}{3} - 0 & 0 - 0 \\ 1 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - \frac{1}{3} & - \frac{4}{3} & - 1 & \frac{1}{3} & 0 \\ 1 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & \frac{1}{3} \end{matrix}] R_{3} = r_{3} - r_{1} [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - \frac{1}{3} & - \frac{4}{3} & - 1 & \frac{1}{3} & 0 \\ 1 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - \frac{1}{3} & - \frac{4}{3} & - 1 & \frac{1}{3} & 0 \\ 1 - 1 & \frac{1}{3} - 1 & \frac{2}{3} - 1 & 0 - 1 & 0 - 0 & \frac{1}{3} - 0 \end{matrix}] R_{2} = - 3 r_{2} [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & - \frac{1}{3} & - \frac{4}{3} & - 1 & \frac{1}{3} & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 (- 3) & - \frac{1}{3} (- 3) & - \frac{4}{3} (- 3) & - 1 (- 3) & \frac{1}{3} (- 3) & 0 (- 3) \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}]$

3Step 3. Calculating Inverse of the function

$R_{1} = r_{1} - r_{2} [\begin{matrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 - 0 & 1 - 1 & 1 - 4 & 1 - 3 & 0 + 1 & 0 - 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] R_{3} = r_{3} + \frac{2}{3} r_{2} [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & - \frac{2}{3} & - \frac{1}{3} & - 1 & 0 & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 + 0 & - \frac{2}{3} + \frac{2}{3} & - \frac{1}{3} + \frac{8}{3} & - 1 + 2 & 0 - \frac{2}{3} & \frac{1}{3} + 0 \end{matrix}] \to [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & \frac{7}{3} & 1 & - \frac{2}{3} & \frac{1}{3} \end{matrix}] R_{3} = r_{3} \cdot \frac{3}{7} [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & \frac{7}{3} & 1 & - \frac{2}{3} & \frac{1}{3} \end{matrix}] \to [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 (\frac{3}{7}) & 0 (\frac{3}{7}) & \frac{7}{3} (\frac{3}{7}) & 1 (\frac{3}{7}) & - \frac{2}{3} (\frac{3}{7}) & \frac{1}{3} (\frac{3}{7}) \end{matrix}] \to [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] R_{1} = r_{1} + 3 r_{3} [\begin{matrix} 1 & 0 & - 3 & - 2 & 1 & 0 \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] \to [\begin{matrix} 1 + 0 & 0 + 0 & - 3 + 3 & - 2 + \frac{9}{7} & 1 - \frac{6}{7} & 0 + \frac{3}{7} \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & - \frac{5}{7} & \frac{1}{7} & \frac{3}{7} \\ 0 & 1 & 4 & 3 & - 1 & 0 \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] R_{1} = r_{2} - 4 r_{3} \to [\begin{matrix} 1 & 0 & 0 & - \frac{5}{7} & \frac{1}{7} & \frac{3}{7} \\ 0 - 0 & 1 - 0 & 4 - 4 & 3 - \frac{12}{7} & - 1 + \frac{8}{7} & 0 - \frac{4}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 & - \frac{5}{7} & \frac{1}{7} & \frac{3}{7} \\ 0 & 1 & 0 & \frac{9}{7} & \frac{1}{7} & - \frac{4}{7} \\ 0 & 0 & 1 & \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] A^{- 1} = [\begin{matrix} - \frac{5}{7} & \frac{1}{7} & \frac{3}{7} \\ \frac{9}{7} & \frac{1}{7} & - \frac{4}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}]$

4Step 4. Solution of the matrix

We know,

$X = A^{- 1} B X = [\begin{matrix} x \\ y \\ z \end{matrix}] = A^{- 1} B = [\begin{matrix} - \frac{5}{7} & \frac{1}{7} & \frac{3}{7} \\ \frac{9}{7} & \frac{1}{7} & - \frac{4}{7} \\ \frac{3}{7} & - \frac{2}{7} & \frac{1}{7} \end{matrix}] \cdot [\begin{array}{l} 9 \\ 8 \\ 1 \end{array}] X = [\begin{array}{l} - \frac{5}{7} (9) + \frac{1}{7} (8) + \frac{3}{7} (1) \\ \frac{9}{7} (9) + \frac{1}{7} (8) - \frac{4}{7} (1) \\ \frac{3}{7} (9) - \frac{2}{7} (8) + \frac{1}{7} (1) \end{array}] = [\begin{matrix} - \frac{45}{7} + \frac{8}{7} + \frac{3}{7} \\ \frac{81}{7} + \frac{8}{7} - \frac{4}{7} \\ \frac{27}{7} - \frac{16}{7} + \frac{1}{7} \end{matrix}] = [\begin{matrix} - \frac{34}{7} \\ \frac{85}{7} \\ \frac{12}{7} \end{matrix}]$

Q. 56

Q. 58

Other exercises in this chapter

Q. 55

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equationsx-y+z=2-2y+z=2-2x-3y=12

View solution

Q. 56

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equation x+2z=2-x+2y+3z=-32x-y=2

View solution

Q. 58

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equation3x+3y+z=8x+2y+z=52x-y+z=4

View solution

Q. 59

In Problems 41– 60, use the inverses found in Problems 31–40 to solve each system of equationsx+y+z=23x+2y-z=733x+y+2z=103

View solution