Question

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

(A) 1
(B) 4
(C) 5
(D) -1

Answer 1

Solve the following system via Gaussian elimination method.

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

The augmented matrix is

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

We use elementary row operations to transform this matrix into a triangular one. We keep the first row and use it to produce all zeros elsewhere in the first column. We have

$$\left( \begin{array}{cccc} 1 & 2 & 3 & |5 \\ 0 & 7 & 7 & |7 \\ 0 & 14 & 16 & |22 \end{array} \right)$$

Next we keep the first and second row and try to have zeros in the second column. And so on. We have

$$\left( \begin{array}{cccc} 1 & 2 & 3 & |5 \\ 0 & 1 & 1 & |1 \\ 0 & 0 & 2 & |8 \end{array} \right)$$ $$\left( \begin{array}{cccc} 1 & 2 & 0 & |-7 \\ 0 & 1 & 0 & |-3 \\ 0 & 0 & 1 & |4 \end{array} \right)$$ $$\left( \begin{array}{cccc} 1 & 0 & 0 & |-1 \\ 0 & 1 & 0 & |-3 \\ 0 & 0 & 1 & |4 \end{array} \right)$$

So, $a = -1$, $b = -3$, $c = 4$

Answer: (D) -1.