Linear Systems Exercise

Linear Systems Exercise

(a) Solve the following system of equations:

$$\begin{cases} -x + 2y = 6 \\ 2x + y = 8 \end{cases}$$

(b) Consider the following system of equations:

$$\begin{cases} 2x_1 – x_2 + 3x_3 = 19 \\ -4x_1 + x_2 + 2x_3 = -12 \\ 3x_1 – 2x_2 – 5x_3 = 1 \end{cases}$$

Use the Gaussian elimination method to determine its solutions, if they exist.

Solutions

(a) Solve the 2×2 system

We have:

$$-x + 2y = 6 \tag{1}$$ $$2x + y = 8 \tag{2}$$

Step 1: Express $x$ in terms of $y$ from equation (1):

$$-x + 2y = 6 \Rightarrow x = 2y – 6$$

Step 2: Substitute into equation (2):

$$2(2y – 6) + y = 8$$ $$4y – 12 + y = 8$$ $$5y = 20$$ $$y = 4$$

Step 3: Substitute back to find $x$:

$$x = 2(4) – 6 = 8 – 6 = 2$$

✅ Solution: $(x, y) = (2, 4)$

(b) Gaussian elimination

The system:

$$\begin{cases} 2x_1 – x_2 + 3x_3 = 19 \\ -4x_1 + x_2 + 2x_3 = -12 \\ 3x_1 – 2x_2 – 5x_3 = 1 \end{cases}$$

Write the augmented matrix:

$$\begin{bmatrix} 2 & -1 & 3 & | & 19 \\ -4 & 1 & 2 & | & -12 \\ 3 & -2 & -5 & | & 1 \end{bmatrix}$$

Step 1: Make first pivot = 1 (Row 1 ÷ 2):

$$\begin{bmatrix} 1 & -0.5 & 1.5 & | & 9.5 \\ -4 & 1 & 2 & | & -12 \\ 3 & -2 & -5 & | & 1 \end{bmatrix}$$

Step 2: Eliminate below pivot

• Row 2: $R_2 + 4 \times R_1$
• Row 3: $R_3 – 3 \times R_1$

Result:

$$\begin{bmatrix} 1 & -0.5 & 1.5 & | & 9.5 \\ 0 & -1 & 8 & | & 26 \\ 0 & -0.5 & -9.5 & | & -27.5 \end{bmatrix}$$

Step 3: Make second pivot = 1 (Row 2 ÷ -1):

$$\begin{bmatrix} 1 & -0.5 & 1.5 & | & 9.5 \\ 0 & 1 & -8 & | & -26 \\ 0 & -0.5 & -9.5 & | & -27.5 \end{bmatrix}$$

Step 4: Eliminate below pivot

• Row 3: $R_3 + 0.5 \times R_2$

Result:

$$\begin{bmatrix} 1 & -0.5 & 1.5 & | & 9.5 \\ 0 & 1 & -8 & | & -26 \\ 0 & 0 & -13.5 & | & -40.5 \end{bmatrix}$$

Step 5: Back substitution

From Row 3:

$$-13.5x_3 = -40.5 \Rightarrow x_3 = \frac{-40.5}{-13.5} = 3$$

From Row 2:

$$x_2 – 8x_3 = -26 \Rightarrow x_2 = -26 + 8 \times 3 = -2$$

From Row 1:

$$x_1 – 0.5x_2 + 1.5x_3 = 9.5$$ $$x_1 – 0.5(-2) + 1.5(3) = 9.5$$ $$x_1 + 1 + 4.5 = 9.5$$ $$x_1 = 9.5 – 5.5 = 4$$

✅ Solution: $(x_1, x_2, x_3) = (4, -2, 3)$

Final Answers

• (a) Solution: $x = 2$, $y = 4$
• (b) Solution: $x_1 = 4$, $x_2 = -2$, $x_3 = 3$