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 c

Answer 1

Answer on Question # 40996 – Math – Linear algebra

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

Solution:

$$\mathbf{A} = \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix}$$ $$\mathbf{B} = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}$$ $$\mathbf{X} = \begin{pmatrix} 1 \\ X \\ 2 \\ X \end{pmatrix}$$ $$\mathbf{A} \cdot \mathbf{X} = \mathbf{B}$$

so

$$\mathbf{X} = \mathbf{A}^{-1} \cdot \mathbf{B}$$

Find the determinant of a matrix $\mathbf{A}$

det $\mathbf{A} = -28$

To find the inverse matrix to calculate the cofactors of the elements of the matrix $\mathbf{A}$

$$C_{1,1} = (-1)^{1+1} \begin{array}{ccc} -1 & -3 & \\ 2 & 1 & \end{array} = 5$$ $$C_{1,2} = (-1)^{1+2} \begin{array}{ccc} 2 & -3 & \\ 5 & 1 & \end{array} = -17$$ $$C_{1,3} = (-1)^{1+3} \begin{array}{ccc} 2 & -1 & \\ 5 & 2 & \end{array} = 9$$ $$C_{2,1} = (-1)^{2+1} \begin{array}{ccc} 3 & 2 & \\ 2 & 1 & \end{array} = 1$$ $$C_{2,2} = (-1)^{2+2} \begin{array}{ccc} 1 & 2 & \\ 5 & 1 & \end{array} = -9$$ $$\mathrm{C}_{2,3} = (-1)^{2+3} \begin{array}{ccc} 1 & 3 & \\ 5 & 2 & = 13 \end{array}$$ $$\mathrm{C}_{3,1} = (-1)^{3+1} \begin{array}{ccc} 3 & 2 & \\ -1 & -3 & = -7 \end{array}$$ $$\mathrm{C}_{3,2} = (-1)^{3+2} \begin{array}{ccc} 1 & 2 & \\ 2 & -3 & = 7 \end{array}$$ $$\mathrm{C}_{3,3} = (-1)^{3+3} \begin{array}{ccc} 1 & 3 & \\ 2 & -1 & = -7 \end{array}$$ $$\mathbf{C} = \begin{pmatrix} 5 & -17 & 9 \\ 1 & -9 & 13 \\ -7 & 7 & -7 \end{pmatrix}$$ $$\mathbf{C}^{\mathrm{T}} = \begin{pmatrix} 5 & 1 & -7 \\ -17 & -9 & 7 \\ 9 & 13 & -7 \end{pmatrix}$$

Let us find the inverse of a matrix

$$\mathbf{A}^{-1} = \begin{array}{l} \mathbf{C}^{\mathrm{T}} \\ \det \mathbf{A} \end{array} = \begin{pmatrix} -5/28 & -1/28 & 1/4 \\ 17/28 & 9/28 & -1/4 \\ -9/28 & -13/28 & 1/4 \end{pmatrix}$$

Aind a solution

$$\mathbf{X} = \mathbf{A}^{-1} \cdot \mathbf{B} = \begin{pmatrix} -5/28 & -1/28 & 1/4 \\ 17/28 & 9/28 & -1/4 \\ -9/28 & -13/28 & 1/4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}$$

Answer: $a = 2$, $b = -3$, $c = 5$.

http://www.AssignmentExpert.com