Question

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

Answer 1

Answer on Question # 40856 – Math – Linear Algebra:

Solve the set of linear equations by Gaussian elimination method:

$$\left\{ \begin{array}{c} a + 2 b + 3 c = 5 \\ 3 a - b + 2 c = 8 \\ 4 a - 6 b - 4 c = - 2 \end{array} \right.$$

Find $c$.

Solution.

Write the system in the matrix form:

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

Lead it to a triangular form. Firstly, subtract the first row, multiplied by 3, from the second row and subtract the first row, multiplied by 4, from the third row:

$$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & -7 & -7 \\ 0 & -14 & -16 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 5 \\ -7 \\ -22 \end{array} \right);$$

Then divide the second row by -7:

$$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & -14 & -16 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 5 \\ 1 \\ -22 \end{array} \right);$$

Now add the second row. Multiplied by 14, to the third row:

$$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 5 \\ 1 \\ -8 \end{array} \right);$$

Lastly, divide the third row by -2:

$$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 5 \\ 1 \\ 4 \end{array} \right);$$

Hence:

$$c = 4 \Rightarrow b = 1 - c = -3 \Rightarrow a = 5 - 2b - 3c = -1 \Rightarrow \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} -1 \\ -3 \\ 4 \end{array} \right).$$

Answer.

$$c = 4.$$