Linear Algebra

The question is asking us to use the Inverse Matrix Method to solve the linear system

$ (a.) \\[3ex] X = A^{-1}B \\[3ex] where \\[3ex] X = \begin{bmatrix} x_1 \\[3ex] x_2 \\[3ex] x_3 \end{bmatrix} \hspace{3em} A = \begin{bmatrix} 2 & -3 & 2 \\[3ex] 3 & 2 & -1 \\[3ex] 1 & -4 & 2 \end{bmatrix} \hspace{3em} B = \begin{bmatrix} 9 \\[3ex] 4 \\[3ex] 6 \end{bmatrix} $

Let us find the inverse of Matrix A
We shall use the Row Reduction Method but you may use the Formula Method if you prefer it.

$ \underline{\text{Row Reduction Method}} \\[3ex] A \ \ | \ \ I = I \ \ | \ \ A^{-1} \\[3ex] \left[ \begin{array}{ccc|ccc} 2 & -3 & 2 & 1 & 0 & 0 \\[3ex] 3 & 2 & -1 & 0 & 1 & 0 \\[3ex] 1 & -4 & 2 & 0 & 0 & 1 \end{array} \right] \begin{matrix} \underrightarrow{-3R_1 + 2R_2} \\ \underrightarrow{-R_1 + 2R_3} \end{matrix} \left[ \begin{array}{ccc|ccc} 2 & -3 & 2 & 1 & 0 & 0 \\[3ex] 0 & 13 & -8 & -3 & 2 & 0 \\[3ex] 0 & -5 & 2 & -1 & 0 & 2 \end{array} \right] \\[10ex] \begin{matrix} \underrightarrow{3R_2 + 13R_1} \\ \underrightarrow{5R_2 + 13R_3} \end{matrix} \left[ \begin{array}{ccc|ccc} 26 & 0 & 2 & 4 & 6 & 0 \\[3ex] 0 & 13 & -8 & -3 & 2 & 0 \\[3ex] 0 & 0 & -14 & -28 & 10 & 26 \end{array} \right] \\[10ex] \begin{matrix} \underrightarrow{R_3 + 7R_1} \\ \underrightarrow{-4R_3 + 7R_2} \end{matrix} \left[ \begin{array}{ccc|ccc} 182 & 0 & 0 & 0 & 52 & 26 \\[3ex] 0 & 91 & 0 & 91 & -26 & -104 \\[3ex] 0 & 0 & -14 & -28 & 10 & 26 \end{array} \right] \\[10ex] \begin{matrix} \underrightarrow{\dfrac{R_1}{182}} \\ \underrightarrow{\dfrac{R_2}{91}} \\ \underrightarrow{-\dfrac{R_3}{14}} \end{matrix} \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & \dfrac{2}{7} & \dfrac{1}{7} \\[5ex] 0 & 1 & 0 & 1 & -\dfrac{2}{7} & -\dfrac{8}{7} \\[5ex] 0 & 0 & 1 & 2 & -\dfrac{5}{7} & -\dfrac{13}{7} \end{array} \right] \\[10ex] \implies \\[3ex] A^{-1} = \begin{bmatrix} 0 & \dfrac{2}{7} & \dfrac{1}{7} \\[5ex] 1 & -\dfrac{2}{7} & -\dfrac{8}{7} \\[5ex] 2 & -\dfrac{5}{7} & -\dfrac{13}{7} \end{bmatrix} $

$ A^{-1} * B \\[3ex] = \begin{bmatrix} 0 & \dfrac{2}{7} & \dfrac{1}{7} \\[5ex] 1 & -\dfrac{2}{7} & -\dfrac{8}{7} \\[5ex] 2 & -\dfrac{5}{7} & -\dfrac{13}{7} \end{bmatrix} \begin{bmatrix} 9 \\[3ex] 4 \\[3ex] 6 \end{bmatrix} \\[16ex] = \begin{bmatrix} 0 * 9 + \dfrac{2}{7} * 4 + \dfrac{1}{7} * 6 \\[5ex] 1 * 9 + -\dfrac{2}{7} * 4 + -\dfrac{8}{7} * 6 \\[5ex] 2 * 9 + -\dfrac{5}{7} * 4 + -\dfrac{13}{7} * 6 \end{bmatrix} \\[16ex] = \begin{bmatrix} 0 + \dfrac{8}{7} + \dfrac{6}{7} \\[5ex] \dfrac{63}{7} - \dfrac{8}{7} - \dfrac{48}{7} \\[5ex] \dfrac{126}{7} - \dfrac{20}{7} - \dfrac{78}{7} \end{bmatrix} \\[16ex] = \begin{bmatrix} \dfrac{14}{7} \\[5ex] \dfrac{7}{7} \\[5ex] \dfrac{28}{7} \end{bmatrix} \\[16ex] = \begin{bmatrix} 2 \\[3ex] 1 \\[3ex] 4 \end{bmatrix} $

$ \begin{bmatrix} x_1 \\[3ex] x_2 \\[3ex] x_3 \end{bmatrix} = \begin{bmatrix} 2 \\[3ex] 1 \\[3ex] 4 \end{bmatrix} \\[10ex] \implies \\[3ex] x_1 = 2 \\[3ex] x_2 = 1 \\[3ex] x_3 = 4 $

Check

$x_1 = 2,\;\;\;x_2 = 1,\;\;\;x_3 = 4$

LHS	RHS
$ 2x_1 - 3x_2 + 2x_3 \\[3ex] 2(2) - 3(1) + 2(4) \\[3ex] 4 - 3 + 8 \\[3ex] 9 $	9
$ 3x_1 + 2x_2 - x_3 \\[3ex] 3(2) + 2(1) - 4 \\[3ex] 6 + 2 - 4 \\[3ex] 4 $	4
$ x_1 - 4x_2 + 2x_3 \\[3ex] 2 - 4(1) + 2(4) \\[3ex] 2 - 4 + 8 \\[3ex] 6 $	6

$ (b.) \\[3ex] \underline{\text{Denominator}} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \hspace{3em} \begin{vmatrix} x + 1 & -5 & -6 \\[3ex] -1 & x & 2 \\[3ex] -3 & 2 & x + 1 \end{vmatrix} = 0 \\[12ex] \underline{\text{1st Row}} \\[3ex] (x + 1)\begin{vmatrix} x & 2 \\[3ex] 2 & x + 1 \end{vmatrix} - -5\begin{vmatrix} -1 & 2 \\[3ex] -3 & x + 1 \end{vmatrix} + -6\begin{vmatrix} -1 & x \\[3ex] -3 & 2 \end{vmatrix} = 0 \\[7ex] (x + 1)[x(x + 1) - 4] + 5[-1(x + 1) + 6] - 6(-2 + 3x) = 0 \\[3ex] (x + 1)(x^2 + x - 4) + 5(-x - 1 + 6) + 12 - 18x = 0 \\[3ex] x^3 + x^2 - 4x + x^2 + x - 4 + 5(-x + 5) + 12 - 18x = 0 \\[3ex] x^3 + 2x^2 - 3x - 4 - 5x + 25 + 12 - 18x = 0 \\[3ex] x^3 + 2x^2 - 26x + 33 = 0 \\[3ex] \underline{\text{Rational Zeros Theorem}} \\[3ex] \text{Leading Coefficient} = 1 \\[3ex] \text{Factors of Leading Coefficient} = \pm 1 \\[3ex] \text{Constant Term} = 33 \\[3ex] \text{Factors of Constant Term} = \pm 1, \pm 3, \pm 11, \pm 33 \\[3ex] \text{Possible Rational Zeros} = \dfrac{\text{Factors of Constant Term}}{\text{Factors of Leading Coefficient}} \\[5ex] = \pm 1, \pm 3, \pm 11, \pm 33 \\[3ex] \text{Try } x = 3 \\[3ex] 3^3 + 2(3)^2 - 26(3) + 33 = 0 \\[3ex] x = 3 \;\;\text{is a solution} \\[3ex] \text{Let us find the remaining two solutions} \\[5ex] \underline{\text{Synthetic Division: By Addition}} \\[3ex] \begin{array}{c|cccc} 3 & 1 & 2 & -26 & 33 \\ + & & 3 & 15 & -33 \\ \hline & 1 & 5 & -11 & 0 \\ \end{array} \\[10ex] Quotient = x^2 + 5x - 11 \\[5ex] \underline{\text{Completing the Square Method}} \\[3ex] \text{coefficient of }x = 5 \\[3ex] \text{half of it} = \dfrac{1}{2} * 5 = \dfrac{5}{2} \\[5ex] \text{square the result} = \left(\dfrac{5}{2}\right)^2 \\[5ex] x^2 + 5x + \left(\dfrac{5}{2}\right)^2 = 11 + \left(\dfrac{5}{2}\right)^2 \\[5ex] \left(x + \dfrac{5}{2}\right)^2 = 11 + \dfrac{25}{4} \\[5ex] \left(x + \dfrac{5}{2}\right)^2 = \dfrac{44}{4} + \dfrac{25}{4} \\[5ex] \left(x + \dfrac{5}{2}\right)^2 = \dfrac{69}{4} \\[5ex] x + \dfrac{5}{2} = \pm \sqrt{\dfrac{69}{4}} = \pm \dfrac{\sqrt{69}}{2} \\[5ex] x = -\dfrac{5}{2} \pm \dfrac{\sqrt{69}}{2} \\[5ex] x = \dfrac{-5 \pm \sqrt{69}}{2} $

(a.)

(b.)