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 a

Answer 1

Answer on Question #58765 – Math – Linear Algebra

Question

Solve the set of linear equations by the matrix method:

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

Solution

$$Ax = y,$$ $$A^{-1}Ax = A^{-1}y,$$ $$x = A^{-1}y,$$ $$A = \begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix}, \quad x = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, \quad y = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix},$$ $$\begin{pmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 3 \\ -8 \\ 9 \end{pmatrix}.$$

The inverse of a matrix $A$ is

$$A^{-1} = \frac{1}{|A|} A_{+}^{T}.$$

The determinant of a matrix $A$ is

$$\begin{aligned} |A| &= \begin{vmatrix} 1 & 3 & 2 \\ 2 & -1 & -3 \\ 5 & 2 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} -1 & -3 \\ 2 & 1 \end{vmatrix} - 3 \begin{vmatrix} 2 & -3 \\ 5 & 1 \end{vmatrix} + 2 \begin{vmatrix} 2 & -1 \\ 5 & 2 \end{vmatrix} = \\ &= 1 \cdot \left(-1 \cdot 1 - 2 \cdot (-3)\right) - 3 \cdot \left(2 \cdot 1 - 5 \cdot (-3)\right) + 2 \cdot \left(2 \cdot 2 - 5 \cdot (-1)\right) = \\ &= -1 + 6 - 3(2 + 15) + 2(4 + 5) = 5 - 51 + 18 = -28. \end{aligned}$$

The minor matrix is

$$M = \begin{pmatrix} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{pmatrix}.$$

The cofactor matrix is

$$A _ {*} = \left( \begin{array}{c c c} M _ {1 1} & - M _ {1 2} & M _ {1 3} \\ - M _ {2 1} & M _ {2 2} & - M _ {2 3} \\ M _ {3 1} & - M _ {3 2} & M _ {3 3} \end{array} \right).$$

Minors are

$$M _ {1 1} = \left| \begin{array}{c c} - 1 & - 3 \\ 2 & 1 \end{array} \right| = - 1 \cdot 1 - 2 \cdot (- 3) = - 1 + 6 = 5,$$ $$M _ {1 2} = \left| \begin{array}{c c} 2 & - 3 \\ 5 & 1 \end{array} \right| = 2 \cdot 1 - 5 \cdot (- 3) = 2 + 1 5 = 1 7,$$ $$M _ {1 3} = \left| \begin{array}{c c} 2 & - 1 \\ 5 & 2 \end{array} \right| = 2 \cdot 2 - 5 \cdot (- 1) = 9,$$ $$M _ {2 1} = \left| \begin{array}{c c} 3 & 2 \\ 2 & 1 \end{array} \right| = 3 \cdot 1 - 2 \cdot 2 = - 1,$$ $$M _ {2 2} = \left| \begin{array}{c c} 1 & 2 \\ 5 & 1 \end{array} \right| = 1 \cdot 1 - 5 \cdot 2 = - 9,$$ $$M _ {2 3} = \left| \begin{array}{c c} 1 & 3 \\ 5 & 2 \end{array} \right| = 1 \cdot 2 - 5 \cdot 3 = - 1 3,$$ $$M _ {3 1} = \left| \begin{array}{c c} 3 & 2 \\ - 1 & - 3 \end{array} \right| = 3 \cdot (- 3) - (- 1) \cdot 2 = - 7,$$ $$M _ {3 2} = \left| \begin{array}{c c} 1 & 2 \\ 2 & - 3 \end{array} \right| = 1 \cdot (- 3) - 2 \cdot 2 = - 7,$$ $$M _ {3 3} = \left| \begin{array}{c c} 1 & 3 \\ 2 & - 1 \end{array} \right| = 1 \cdot (- 1) - 2 \cdot 3 = - 7.$$

The cofactor matrix will be

$$A _ {*} = \left( \begin{array}{c c c} 5 & - 1 7 & 9 \\ 1 & - 9 & 1 3 \\ - 7 & 7 & - 7 \end{array} \right).$$

The transpose of matrix $A_{*}$ is

$$A _ {*} ^ {T} = \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right).$$

The inverse of a matrix $A$ is

$$A ^ {- 1} = - \frac {1}{2 8} \cdot \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right).$$

The solution of the system is

$$\begin{array}{l} x = \left( \begin{array}{c} a \\ b \\ c \end{array} \right) = \left( \begin{array}{c c c} - \frac {5}{2 8} & - \frac {1}{2 8} & \frac {1}{4} \\ \frac {1 7}{2 8} & \frac {9}{2 8} & - \frac {1}{4} \\ - \frac {9}{2 8} & - \frac {1 3}{2 8} & \frac {1}{4} \end{array} \right) \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) = \left( \begin{array}{c} \frac {- 5 \cdot 3}{2 8} + \frac {- 1 \cdot (- 8)}{2 8} + \frac {1 \cdot 9}{4} \\ \frac {1 7 \cdot 3}{2 8} + \frac {9 \cdot (- 8)}{2 8} + \frac {(- 1) \cdot 9}{4} \\ - \frac {9 \cdot 3}{2 8} + \frac {- 1 3 \cdot (- 8)}{2 8} + \frac {1 \cdot 9}{2 8} \end{array} \right) \\ = \left( \begin{array}{c} - \frac {1 5}{2 8} + \frac {8}{2 8} + \frac {9}{4} \\ \frac {5 1}{2 8} - \frac {7 2}{2 8} - \frac {9}{4} \\ - \frac {2 7}{2 8} + \frac {1 0 4}{2 8} + \frac {9}{4} \end{array} \right) = \\ = \left( \begin{array}{c} \frac {- 1 5 + 8 + 9 \cdot 7}{2 8} \\ \frac {5 1 - 7 2 - 9 \cdot 7}{2 8} \\ \frac {- 2 7 + 1 0 4 + 7 \cdot 9}{2 8} \end{array} \right) = \left( \begin{array}{c} \frac {5 6}{2 8} \\ \frac {- 8 4}{2 8} \\ \frac {1 4 0}{2 8} \end{array} \right) = \left( \begin{array}{c} 2 \\ - 3 \\ 5 \end{array} \right), \\ \end{array}$$

that is, $a = 2, b = -3, c = 5$ .

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

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