Question

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

Answer 1

Answer Question #40857 – Math - Linear Algebra

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

Solution:

$$\left\{ \begin{array}{l} a + 3b + 2c = 3 \\ 2a - b - 3c = -8 \\ 5a + 2b + c = 9 \end{array} \right.$$

By matrix method:

$$A * x = b \quad => \quad \vec{x} = A^{-1} * \vec{b}$$ $$\left( \begin{array}{ccc} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{array} \right) \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right) \quad => \quad A * \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right)$$

Let's find $A^{-1}$:

$$A^{-1} = \frac{1}{28} \left( \begin{array}{ccc} -5 & -1 & 7 \\ 17 & 9 & -7 \\ -9 & -13 & 7 \end{array} \right)$$

From this:

$$\left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \frac{1}{28} \left( \begin{array}{ccc} -5 & -1 & 7 \\ 17 & 9 & -7 \\ -9 & -13 & 7 \end{array} \right) * \left( \begin{array}{c} 3 \\ -8 \\ 9 \end{array} \right) = \left( \begin{array}{c} 2 \\ -3 \\ 5 \end{array} \right)$$

So:

$$a = 2, b = -3, c = 5$$

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