Linear Algebra

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These are the solutions to the CSEC past questions on the topics: Linear Algebra.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided for some questions.
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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} $

Formula Sheet: List of Formulae

(1.) Determine the values of the unknowns in EACH of the matrix equations below.

$ (i.) \begin{bmatrix} 4 & 0 \\[2ex] -2 & 5 \end{bmatrix} + \begin{bmatrix} x & 2 \\[2ex] 8 & -1 \end{bmatrix} = \begin{bmatrix} -3 & 2 \\[2ex] y & 4 \end{bmatrix} \\[10ex] (ii.) \begin{bmatrix} 5 & -3 \\[2ex] 2 & 3 \end{bmatrix} \begin{bmatrix} a & 2 \\[2ex] c & -1 \end{bmatrix} = \begin{bmatrix} -10 & 13 \\[2ex] 17 & 1 \end{bmatrix} $

$ (i.) \\[3ex] 4 + x = -3 \\[3ex] x = -3 - 4 \\[3ex] x = -7 \\[5ex] -2 + 8 = y \\[3ex] y = 6 \\[5ex] (ii.) \\[3ex] \begin{bmatrix} 5 & -3 \\[2ex] 2 & 3 \end{bmatrix} \begin{bmatrix} a & 2 \\[2ex] c & -1 \end{bmatrix} \\[7ex] \begin{bmatrix} 5a + -3c & 5(2) + (-3)(-1) \\[2ex] 2a + 3c & 2(2) + 3(-1) \end{bmatrix} \\[7ex] \begin{bmatrix} 5a - 3c & 10 + 3 \\[2ex] 2a + 3c & 4 - 3 \end{bmatrix} \\[7ex] \begin{bmatrix} 5a - 3c & 13 \\[2ex] 2a + 3c & 1 \end{bmatrix} \\[7ex] So: \\[3ex] \begin{bmatrix} 5a - 3c & 13 \\[2ex] 2a + 3c & 1 \end{bmatrix} = \begin{bmatrix} -10 & 13 \\[2ex] 17 & 1 \end{bmatrix} \\[5ex] $ This implies that:

$ 5a - 3c = -10...eqn.(1) \\[3ex] 2a + 3c = 17 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] 7a = 7 \\[3ex] a = \dfrac{7}{7} \\[5ex] a = 1 \\[5ex] \text{Substitute a = 1 in eqn.(2)} \\[3ex] 2(1) + 3c = 17 \\[3ex] 2 + 3c = 17 \\[3ex] 3c = 17 - 2 \\[3ex] 3c = 15 \\[3ex] c = \dfrac{15}{3} \\[5ex] c = 5 $

(2.) The determinant of the matrix $\begin{bmatrix} 6 & 2v \\[2ex] -5 & -v \end{bmatrix}$ is 24.
Calculate the value of v

$ -6v - (-10v) = 24 \\[3ex] -6v + 10v = 24 \\[3ex] 4v = 24 \\[3ex] v = \dfrac{24}{4} \\[5ex] v = 6 $

$ \begin{bmatrix} 6 & 2v \\[2ex] -5 & -v \end{bmatrix} = \begin{bmatrix} 6 & 2(6) \\[2ex] -5 & -6 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\[2ex] -5 & -6 \end{bmatrix} $

Calculator2-1st

(3.) The matrices L and M are defined as follows.

$ L = \begin{bmatrix} 9 & 5 \\[2ex] 3 & 2 \end{bmatrix}, \hspace{3em} M = \begin{bmatrix} 2 \\[2ex] -4 \end{bmatrix} $

Evaluate EACH of the following.
(i.) The matrix product LM

(ii.) $L^{-1}$, the inverse of L

$ (i.) \\[3ex] LM \\[3ex] = \begin{bmatrix} 9 & 5 \\[2ex] 3 & 2 \end{bmatrix} \begin{bmatrix} 2 \\[2ex] -4 \end{bmatrix} \\[8ex] = \begin{bmatrix} 9(2) + 5(-4) \\[2ex] 3(2) + 2(-4) \end{bmatrix} \\[8ex] = \begin{bmatrix} 18 -20 \\[2ex] 6 - 8 \end{bmatrix} \\[8ex] = \begin{bmatrix} -2 \\[2ex] -2 \end{bmatrix} \\[5ex] $ We have at least two approaches of determining the inverse of a matrix.
Use any approach you prefer (or required by your teacher).

$ (ii.) \\[3ex] \underline{\text{1st Approach: By Formula}} \\[3ex] L = \begin{bmatrix} 9 & 5 \\[2ex] 3 & 2 \end{bmatrix} \\[8ex] adj\;L = \begin{bmatrix} 2 & -5 \\[2ex] -3 & 9 \end{bmatrix} \\[8ex] det\;L = \begin{vmatrix} 9 & 5 \\[3ex] 3 & 2 \end{vmatrix} \\[8ex] = 9(2) - 3(5) \\[3ex] = 18 - 15 \\[3ex] = 3 \\[5ex] L^{-1} = \dfrac{adj\;L}{det\;L} \\[5ex] = \dfrac{\begin{bmatrix} 2 & -5 \\[2ex] -3 & 9 \end{bmatrix}}{3} \\[7ex] = \begin{bmatrix} \dfrac{2}{3} & -\dfrac{5}{3} \\[5ex] -\dfrac{3}{3} & \dfrac{9}{3} \end{bmatrix} \\[12ex] = \begin{bmatrix} \dfrac{2}{3} & -\dfrac{5}{3} \\[5ex] -1 & 3 \end{bmatrix} $

$ \underline{\text{2nd Approach: Row Reduction Method}} \\[3ex] L \ \ | \ \ I = I \ \ | \ \ L^{-1} \\[3ex] \left[ \begin{array}{cc|cc} 9 & 5 & 1 & 0 \\[3ex] 3 & 2 & 0 & 1 \end{array} \right] \underrightarrow{-R_1 + 3R_2} \left[ \begin{array}{cc|cc} 9 & 5 & 1 & 0 \\[3ex] 0 & 1 & -1 & 3 \end{array} \right] \\[10ex] \underrightarrow{-5R_2 + R_1} \left[ \begin{array}{cc|cc} 9 & 0 & 6 & -15 \\[3ex] 0 & 1 & -1 & 3 \end{array} \right] \\[10ex] \underrightarrow{R_1 \div 9} \left[ \begin{array}{cc|cc} 1 & 0 & \dfrac{6}{9} & -\dfrac{15}{9} \\[5ex] 0 & 1 & -1 & 3 \end{array} \right] \\[10ex] \implies \\[3ex] L^{-1} = \begin{bmatrix} \dfrac{6}{9} & -\dfrac{15}{9} \\[5ex] -1 & 3 \end{bmatrix} = \begin{bmatrix} \dfrac{2}{3} & -\dfrac{5}{3} \\[5ex] -1 & 3 \end{bmatrix} $

(7.) (a.) The matrices P, Q and R are given below, in terms of the scalar constants a, b and c, as

$ P = \begin{bmatrix} 3 & -9 \\[2ex] a & 7 \end{bmatrix}, \hspace{2em} Q = \begin{bmatrix} -1 & b \\[2ex] -4 & 1 \end{bmatrix}, \hspace{2em} R = \begin{bmatrix} c & -3 \\[2ex] -4 & 8 \end{bmatrix} $

Given that P + Q = R, find the value a, b and c

(b.) Solve the following pair of simultaneous equations using a matrix method.

$ 5x - 2y = 44 \\[3ex] 2x + 3y = 10 \\[3ex] $

$ (a.) \\[3ex] 3 + (-1) = c \\[3ex] c = 3 - 1 \\[3ex] c = 2 \\[5ex] -9 + b = -3 \\[3ex] b = -3 + 9 \\[3ex] b = 6 \\[5ex] a + (-4) = -4 \\[3ex] a - 4 = -4 \\[3ex] a = -4 + 4 \\[3ex] a = 0 \\[3ex] $ (b.) Using a matrix method probably means using any of these methods:
Method of Determinants (also known as Cramer's Rule)
Row Reduction Method (also known as the Guass-Jordan Method or the Guassian Elimination Method)
Matrix Inverse Method
We shall solve the simultaneous equation using those three methods/approaches.
Then, we shall check our solution.
Use any method required by your teacher.

$ \underline{\text{1st Approach: Cramer's Rule}} \\[3ex] \begin{bmatrix} 5 & -2 \\[2ex] 2 & 3 \end{bmatrix} \begin{bmatrix} x \\[2ex] y \end{bmatrix} = \begin{bmatrix} 44 \\[2ex] 10 \end{bmatrix} \\[10ex] \underline{\text{Denominator}} \\[3ex] \begin{vmatrix} 5 & -2 \\[3ex] 2 & 3 \end{vmatrix} \\[7ex] = 5(3) - 2(-2) \\[3ex] = 15 + 4 \\[3ex] = 19 \\[5ex] \underline{\text{Numerator for }x} \\[3ex] \begin{vmatrix} 44 & -2 \\[3ex] 10 & 3 \end{vmatrix} \\[7ex] = 44(3) - 10(-2) \\[3ex] = 132 + 20 \\[3ex] = 152 \\[5ex] x = \dfrac{152}{19} = 8 \\[5ex] \underline{\text{Numerator for }y} \\[3ex] \begin{vmatrix} 5 & 44 \\[3ex] 2 & 10 \end{vmatrix} \\[7ex] = 5(10) - 2(44) \\[3ex] = 50 - 88 \\[3ex] = -38 \\[5ex] y = \dfrac{-38}{19} = -2 $

$ \underline{\text{2nd Approach: Gauss-Jordan Method}} \\[3ex] \left[ \begin{array}{cc|c} 1 & 0 & x \\[3ex] 0 & 1 & y \end{array} \right] \\[10ex] \left[ \begin{array}{cc|c} 5 & -2 & 44 \\[3ex] 2 & 3 & 10 \end{array} \right] \underrightarrow{-2R_1 + 5R_2} \left[ \begin{array}{cc|c} 5 & -2 & 44 \\[3ex] 0 & 19 & -38 \end{array} \right] \underrightarrow{2R_2 + 19R_1} $

$ \left[ \begin{array}{cc|c} 95 & 0 & 760 \\[3ex] 0 & 19 & -38 \end{array} \right] \begin{matrix} \underrightarrow{R_1 \div 95} \\ \underrightarrow{R_2 \div 19} \end{matrix} \left[ \begin{array}{cc|c} 1 & 0 & 8 \\[3ex] 0 & 1 & -2 \end{array} \right] \\[10ex] x = 8 \\[3ex] y = -2 $

$ \underline{\text{3rd Approach: Matrix Inverse Method}} \\[3ex] Let\;\;A = \begin{bmatrix} 5 & -2 \\[2ex] 2 & 3 \end{bmatrix} \hspace{3em} B = \begin{bmatrix} 44 \\[2ex] 10 \end{bmatrix} \\[10ex] adj\;A = \begin{bmatrix} 3 & -(-2) \\[2ex] -2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\[2ex] -2 & 5 \end{bmatrix} \\[10ex] det\;A = \begin{vmatrix} 5 & -2 \\[3ex] 2 & 3 \end{vmatrix} \\[7ex] = 5(3) - 2(-2) \\[3ex] = 15 + 4 \\[3ex] = 19 \\[5ex] A^{-1} = \dfrac{adj\;A}{det\;A} \\[5ex] = \dfrac{\begin{bmatrix} 3 & 2 \\[2ex] -2 & 5 \end{bmatrix}}{19} \\[10ex] = \begin{bmatrix} \dfrac{3}{19} & \dfrac{2}{19} \\[5ex] -\dfrac{2}{19} & \dfrac{5}{19} \end{bmatrix} \\[12ex] \begin{bmatrix} x \\[2ex] y \end{bmatrix} = A^{-1} * B \\[10ex] = \begin{bmatrix} \dfrac{3}{19} & \dfrac{2}{19} \\[5ex] -\dfrac{2}{19} & \dfrac{5}{19} \end{bmatrix} * \begin{bmatrix} 44 \\[2ex] 10 \end{bmatrix} \\[12ex] = \begin{bmatrix} \dfrac{3}{19}(44) & \dfrac{2}{19}(10) \\[5ex] -\dfrac{2}{19}(44) & \dfrac{5}{19}(10) \end{bmatrix} \\[12ex] = \begin{bmatrix} \dfrac{132}{19} + \dfrac{20}{19} \\[5ex] -\dfrac{88}{19} + \dfrac{50}{19} \end{bmatrix} \\[12ex] = \begin{bmatrix} \dfrac{152}{19} \\[5ex] -\dfrac{38}{19} \end{bmatrix} \\[12ex] = \begin{bmatrix} 8 \\[2ex] -2 \end{bmatrix} \\[10ex] x = 8 \\[3ex] y = -2 \\[3ex] $ Check

$x = 8, \;\;\; y = -2$
LHS	RHS
$ 5x - 2y \\[3ex] 5(8) - 2(-2) \\[3ex] 40 + 4 \\[3ex] 44 $	$44$
$ 2x + 3y \\[3ex] 2(8) + 3(-2) \\[3ex] 16 - 6 \\[3ex] 10 $	$10$

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