Question

Solve the following system of linear equations by inversion method:
x1 + 6x2 + 4x3 = 2
2x1 + 4x2 - x3 = 3
-x1 + 3x2 + 5x3 =3

Answer 1

Conditions

Solve the following system of linear equations by inversion method:

$$x1 + 6x2 + 4x3 = 2$$ $$2x1 + 4x2 - x3 = 3$$ $$-x1 + 3x2 + 5x3 = 3$$

Solution

Let's write the system in a matrix form:

$$AX = B: \begin{pmatrix} 1 & 6 & 4 \\ 2 & 4 & -3 \\ -1 & 3 & 5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 3 \end{pmatrix}$$

Let's find the inversion matrix $A^{-1}$ by using cofactors method:

$$A^{-1} = \frac{1}{detA} \begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix}, \quad A_{ij} = (-1)^{i+j} M_{ij}$$ $$A_{11} = (-1)^2 \begin{vmatrix} 4 & -3 \\ 3 & 5 \end{vmatrix} = 20 + 9 = 29$$

...

$$A_{33} = (-1)^6 \begin{vmatrix} 1 & 6 \\ 2 & 4 \end{vmatrix} = 4 - 12 = -8$$ $$\begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix} = \begin{pmatrix} 29 & -18 & -34 \\ -7 & 9 & 11 \\ 10 & -9 & -8 \end{pmatrix}$$ $$\det(A) = \begin{vmatrix} 1 & 6 & 4 \\ 2 & 4 & -3 \\ -1 & 3 & 5 \end{vmatrix} = 20 + 9 - 60 + 16 + 18 + 24 = 27$$ $$A^{-1} = \frac{1}{27} \begin{pmatrix} 29 & -18 & -34 \\ -7 & 9 & 11 \\ 10 & -9 & -8 \end{pmatrix}$$

Now let's multiply our matrix equation from the left side by inversion matrix:

$$A^{-1}AX = A^{-1}B$$ $$X = A^{-1}B$$ $$X = \frac{1}{27} \begin{pmatrix} 29 & -18 & -34 \\ -7 & 9 & 11 \\ 10 & -9 & -8 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \\ 3 \end{pmatrix} = \frac{1}{27} \begin{pmatrix} 29 \cdot 2 - 18 \cdot 3 - 34 \cdot 3 \\ -7 \cdot 2 + 9 \cdot 3 + 11 \cdot 3 \\ 10 \cdot 2 - 9 \cdot 3 - 8 \cdot 3 \end{pmatrix}$$ $$X = \left( \begin{array}{c} -98 \\ \hline 27 \\ \frac{46}{27} \\ \frac{-31}{27} \end{array} \right)$$