Question

Solve the set of linear equations by Gaussian elimination method: a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find b.

(A) 4

(B) -5

(C) -3

(D) 5

Answer 1

The Gauss method.

We write the system in the form:

$$\left( \begin{array}{cc} 1 & 2 & 3 \\ 3 & -1 & 2 \\ 4 & -6 & -4 \end{array} \right) = \left( \begin{array}{c} 5 \\ 8 \\ -2 \end{array} \right)$$

We multiply the first row in (3). Multiplying the second row by (-1). Add the second line to the first:

$$\left( \begin{array}{ccc} 0 & 7 & 7 \\ 3 & -1 & 2 \\ 4 & -6 & -4 \end{array} \right) = \left( \begin{array}{c} 7 \\ 8 \\ -2 \end{array} \right)$$

Multiplying the second row in (4). Multiplying the 3-th row in (3). Add the 3rd row to the second:

$$\left( \begin{array}{ccc} 0 & 7 & 7 \\ 0 & 14 & 20 \\ 4 & -6 & -4 \end{array} \right) = \left( \begin{array}{c} 7 \\ 38 \\ -2 \end{array} \right)$$

We multiply the first row in (2). Multiplying the second row by (-1). Add the second line to the first:

$$\left( \begin{array}{cc} 0 & 0 & -6 \\ 0 & 14 & 20 \\ 4 & -6 & -4 \end{array} \right) = \left( \begin{array}{c} -24 \\ 38 \\ -2 \end{array} \right)$$

From the first line express x3 --- c

$$x_3 = \frac{-24}{-6} = 4$$

From the second row express x2 --- b

$$x_2 = \frac{38 - 20 \cdot 4}{14} = \frac{-42}{14} = -3$$

From the 3rd row express x1 --- a

$$x_1 = \frac{-2 - (-6) \cdot (-3) - (-4) \cdot 4}{4} = \frac{-4}{4} = -1$$ $$b = -3$$