# Question 9.8: Solve the following equations by Gaussian elimination: 2x +......

Solve the following equations by Gaussian elimination:

2x + y + z = 12
6x + 5y − 3z = 6
4x − y + 3z = 5

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.

The equations are already written in the required form: variables in order on the LHS and constants on the RHS. Therefore, write down the augmented matrix and proceed with the Gaussian elimination:

$\begin{matrix} \\ 2x + y + z = 12 \\ 6x + 5y − 3z = 6 \\ 4x − y + 3z = 5 \end{matrix}$  →    $\begin{matrix} \text {x y z RHS} \\ \left(\begin{matrix} 2 & 1 & 1& 12 \\ 6 & 5 & -3 & 6 \\ 4 & -1 & 3 & 5 \end{matrix} \right) \end{matrix}$

Now proceed with the elimination on the augmented matrix.

The elimination is now complete; solve by back substitution. From equation (3)²

−8z = −64      hence         z = $\frac {−64}{-8} = 8$

Substitute z = 8 into equation (2)²

y − (3 × 8) = −15      hence    y = −15 + (3 × 8) = 9

Substitute z = 8 and y = 9 into equation (1)¹

x + (0.5 × 9) + (0.5 × 8) = 6    hence      x = 6 − 4.5 − 4 = −2.5

x=−2.5, y = 9, z = 8

 Action Augmented matrix Calculations Make the boxed 2 into a 1 by dividing row 1 by 2 $\begin{matrix} \text {x y z RHS} \\ \left(\begin{matrix} \fbox{2} & 1 & 1 & 12 \\ 6 & 5 & -3 & 6 \\ 4 & -1 & 3 & 5 \end{matrix} \right) \end{matrix} \begin{matrix} \\(1) \\(2) \\(3) \end{matrix}$ $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 6 & 5 & -3 & 6 \\ 4 & -1 & 3 & 5 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2) \\(3) \end{matrix}$ row 2 + (6 × row 1¹) $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 0 & 2 & -6 & -30 \\ 4 & -1 & 3 & 5 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2)¹ \\(3) \end{matrix}$ $\left\{\begin{array}{r r r}{{-6}}&{{-3}}&{{-3}}&{{-36}}\\ {{6}}&{{5}}&{{-3}}&{{6}}\\ \hline \\ {{0}}&{{2}}&{{-6}}&{{30}}\end{array}\right. \begin{matrix} (1)¹×−6\\ (2) \\ \\ adding \end{matrix}$ row 3 + (–4 × row 1¹) $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 0 & 2 & -6 & -30 \\ 0 & -3 & 1 & -19 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2)¹ \\(3)¹ \end{matrix}$ $\left\{\begin{array}{r r r}{{-4}}&{{-2}}&{{-2}}&{{-24}}\\ {{4}}&{{-1}}&{{-3}}&{{5}}\\ \hline \\ {{0}}&{{-3}}&{{1}}&{{-19}}\end{array}\right. \begin{matrix} (1)¹×−4\\ (3) \\ \\ adding \end{matrix}$ Make the boxed 2 into a 1 by dividing row 2¹ by 2 $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 0 & \fbox{2} & -6 & -30 \\ 0 & -3 & 1 & -19 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2)¹ \\(3)¹ \end{matrix}$ $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 0 & 1 & -3 & -15 \\ 0 & -3 & 1 & -19 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2)² \\(3)¹ \end{matrix}$ row 3¹ + (−3 × row 2²) $\begin{matrix} \left(\begin{matrix} 1 & 0.5 & 0.5 & 6 \\ 0 & 1 & -3 & -15 \\ 0 & 0 & -8 & -64 \end{matrix} \right) \end{matrix} \begin{matrix} (1)¹ \\(2)² \\(3)¹ \end{matrix}$ $\left\{\begin{array}{r r r}{{0}}&{{-3}}&{{-9}}&{{-45}}\\ {{0}}&{{-3}}&{{1}}&{{-19}}\\ \hline \\ {{0}}&{{-0}}&{{-8}}&{{-64}}\end{array}\right. \begin{matrix} (2)²×3\\ (3)¹ \\ \\ adding \end{matrix}$

Question: 9.7

## Solve the following equations by Gaussian elimination: x + y − z = 3 (1) 2x + y − z = 4 (2) 2x + 2y + z = 12 (3) ...

All three equations must be written in the same fo...
Question: 9.20

## Given the input/output table for the three-sector economy: ...

Step 1: Use the underlying assumption total input=...
Question: 9.18

## (a) Find the inverse of the matrix D = (1 0 -2 2 2 3 1 3 2) (b) Show that DD^0−1 = I. ...

(a) Use the definition of the inverse given in equ...
Question: 9.17

## (a) Find the inverse of the matrix D = ( 1 0 -2 2 2 3 1 3 2) by Gauss–Jordan elimination. (b) Show that DD^–1 = I. ...

(a) Write out the augmented matrix consisting of t...
Question: 9.12

## Given the supply and demand functions for two related goods, A and B,Good A :{Qda = 30 − 8Pa + 2Pb {Qsa = −15 + 7Pa Good B :{Qdb = 28 + 4Pa − 6Pb {Qsb = 12 + 2Pb (a) Write down the equilibrium condition for each good. Hence, deduce two equations in Pa and Pb . (b) Use Cramer’s rule to find the ...

(a) The equilibrium condition for each good is tha...
Question: 9.9

## Solve the following equations by Gauss–Jordan elimination: 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 ...

Rearrange the equations to have variables on the L...
Question: 9.2

## A company manufactures two types of wrought iron gates. The number of labour-hours required to produce each type of gate, along with the maximum number of hours available, are given in Table 9.3. ...

(a) For x type I gates and y type II gates, the nu...
Question: 9.21

## (a) Find the inverse of A = (2 1 1 6 5 -3 4 -1 3) by the elimination method, hence solve the equations 2x + y + z = 12 6x + 5y − 3z = 6 4x − y + 3z = 5 (b) Solve the equations in (a) by Gauss–Jordan elimination. ...

(a) Set up the augmented matrix, consisting of the...
Question: 9.19