Linear Algebra
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Linear Transformation to Matrix
For a linear transformation $T:(x, y) \rightarrow (ax + by, cx + dy)$,
the corresponding matrix is:
$
T = \begin{bmatrix}
a & b \\[3ex]
c & d
\end{bmatrix}
$
2 by 2 Matrix
Let:
$
A = \begin{bmatrix}
a_{11} & a_{12} \\[3ex]
a_{21} & a_{22}
\end{bmatrix} = \begin{bmatrix}
\color{red}{a} & \color{darkblue}{b} \\[3ex]
\color{darkblue}{c} & \color{red}{d}
\end{bmatrix} \\[10ex]
(1.)\;\; minor\:\: A = \begin{bmatrix}
d & c \\[3ex]
b & a
\end{bmatrix} \\[10ex]
\text{The signs to remember cofactors when determining the determinant of a matrix is}: \\[3ex]
\begin{vmatrix}
+ & - \\[3ex]
- & +
\end{vmatrix} \\[10ex]
(2.)\;\; cofactor\:\: A = \begin{bmatrix}
d & -c \\[3ex]
-b & a
\end{bmatrix} \\[10ex]
(3.)\;\; adj\:\: A = \begin{bmatrix}
d & -b \\[3ex]
-c & a
\end{bmatrix} \\[10ex]
(4.)\;\; det\;A = ad - cb \\[3ex]
(5.)\;\; A^{-1} = \dfrac{adj\;A}{det\;A}
$
3 by 3 Matrix
Let:
$
A = \begin{bmatrix}
a_{11} & a_{12} & a_{13} \\[3ex]
a_{21} & a_{22} & a_{23} \\[3ex]
a_{31} & a_{32} & a_{33}
\end{bmatrix} =
\begin{bmatrix}
a & b & c \\[3ex]
d & e & f \\[3ex]
g & h & i
\end{bmatrix} \\[15ex]
(1.)\;\; minor\:\: A = \begin{bmatrix}
ei - hf & di - gf & dh - ge \\[3ex]
bi - hc & ai - gc & ah - gb \\[3ex]
bf - ec & af - dc & ae - db
\end{bmatrix} \\[15ex]
\text{The signs to remember cofactors when determining the determinant of a matrix is}: \\[3ex]
\begin{vmatrix}
+ & - & + \\[3ex]
- & + & - \\[3ex]
+ & - & +
\end{vmatrix} \\[10ex]
(2.)\;\; cofactor\:\: A = \begin{bmatrix}
ei - hf & gf - di & dh - ge \\[3ex]
hc - bi & ai - gc & gb - ah \\[3ex]
bf - ec & dc - af & ae - db
\end{bmatrix} \\[15ex]
(3.)\;\; adj\: A = \begin{bmatrix}
ei - hf & hc - bi & bf - ec \\[3ex]
gf - di & ai - gc & dc - af \\[3ex]
dh - ge & gb - ah & ae - db
\end{bmatrix} \\[15ex]
(4.)\;\; det\: A = aei + bgf + cdh - ahf - bdi - cge \\[3ex]
(5.)\;\; A^{-1} = \dfrac{adj\;A}{det\;A}
$
$ 2x_1 - 3x_2 + 2x_3 = 9 \\[3ex] 3x_1 + 2x_2 - x_3 = 4 \\[3ex] x_1 - 4x_2 + 2x_3 = 6 \\[3ex] $ (b.) Solve for x
$ \begin{vmatrix} x + 1 & -5 & -6 \\[3ex] -1 & x & 2 \\[3ex] -3 & 2 & x + 1 \end{vmatrix} \\[10ex] $
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
| 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.)
