Question

Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Solve for a

Answer 1

Answer on question 35904 – Math – Linear Algebra

Solve the set of linear equations by the matrix method: $a + 3b + 2c = 3$, $2a - b - 3c = -8$, $5a + 2b + c = 9$. Solve for a

Solution

a, b and c are variables. So we get this system in the matrix form

$$\left( \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right).$$

Gaussian elimination method gives us

$$\begin{array}{l} \left( \begin{array}{rrr|r} 1 & 3 & 2 & 3 \\ 2 & -1 & -3 & -8 \\ 5 & 2 & 1 & 9 \end{array} \right) \sim \left( \begin{array}{c} (1) \\ (2) - 2(1) \\ (3) - 5(1) \end{array} \right) \sim \left( \begin{array}{rrr|r} 1 & 3 & 2 & 3 \\ 0 & -7 & -7 & -14 \\ 0 & -13 & -9 & -6 \end{array} \right) \sim \left( \begin{array}{c} (1) \\ (2) \\ -7 \\ -(3) \end{array} \right) \\ \sim \left( \begin{array}{rrr|r} 1 & 3 & 2 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 13 & 9 & 6 \end{array} \right) \sim \left( \begin{array}{c} (1) \\ (2) \\ (3) - 13(2) \end{array} \right) \sim \left( \begin{array}{rrr|r} 1 & 3 & 2 & 3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 4 & 20 \end{array} \right) \sim \left( \begin{array}{c} (1) + 2(3) \\ (2) - (3)/4 \\ \frac{(3)}{4} \end{array} \right) \\ \sim \left( \begin{array}{rrr|r} 1 & 3 & 0 & -7 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 5 \end{array} \right) \sim \left( \begin{array}{c} (1) - 3(2) \\ (2) \\ (3) \end{array} \right) \sim \left( \begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & -5 \end{array} \right). \end{array}$$

Therefore, $a = 2$.

Answer: 2.