Linear Algebra
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These are the solutions to the WASSCE Further Mathematics past questions on 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} \\[5ex]
$
(6.) If transformation $A$ is followed by transformation $B$, the combined matrix is $B \times A$.
$ \begin{vmatrix} 3 & 5 & -4 \\[3ex] 6 & -3 & -5 \\[3ex] -2 & 2 & 1 \end{vmatrix} \\[10ex] $ (b.) Using your result in (a.), solve the simultaneous equation
$ 3x + 5y - 4z = 1 \\[3ex] 6x - 3y - 5z = -15 \\[3ex] -2x + 2y + z = 5 \\[3ex] $
The question is asking us to use the Method of Determinants (also known as Cramer's Rule)
$ \begin{bmatrix} 3 & 5 & -4 \\[3ex] 6 & -3 & -5 \\[3ex] -2 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \\[3ex] z \end{bmatrix} = \:\:\: \begin{bmatrix} 1 \\[3ex] -15 \\[3ex] 5 \end{bmatrix} \\[15ex] (a.) \\[3ex] \underline{\text{Denominator}} \\[3ex] \begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \hspace{3em} \begin{vmatrix} 3 & 5 & -4 \\[3ex] 6 & -3 & -5 \\[3ex] -2 & 2 & 1 \end{vmatrix} \\[12ex] \underline{\text{1st Row}} \\[3ex] = 3\begin{vmatrix} -3 & -5 \\[3ex] 2 & 1 \end{vmatrix} -5\begin{vmatrix} 6 & -5 \\[3ex] -2 & 1 \end{vmatrix} + -4\begin{vmatrix} 6 & -3 \\[3ex] -2 & 2 \end{vmatrix} \\[12ex] = 3(-3 + 10) - 5(6 - 10) - 4(12 - 6) \\[3ex] = 3(7) - 5(-4) - 4(6) \\[3ex] = 21 + 20 - 24 \\[3ex] = 17 $
$ \underline{\text{Numerator for }x} \\[3ex] \begin{vmatrix} 1 & 5 & -4 \\[3ex] -15 & -3 & -5 \\[3ex] 5 & 2 & 1 \end{vmatrix} \\[12ex] \underline{\text{1st Row}} \\[3ex] = 1\begin{vmatrix} -3 & -5 \\[3ex] 2 & 1 \end{vmatrix} -5\begin{vmatrix} -15 & -5 \\[3ex] 5 & 1 \end{vmatrix} + -4\begin{vmatrix} -15 & -3 \\[3ex] 5 & 2 \end{vmatrix} \\[10ex] = (-3 + 10) - 5(-15 + 25) - 4(-30 + 15) \\[3ex] = 7 - 5(10) - 4(-15) \\[3ex] = 7 - 50 + 60 \\[3ex] = 17 $
$ \underline{\text{Numerator for }y} \\[3ex] \begin{vmatrix} 3 & 1 & -4 \\[3ex] 6 & -15 & -5 \\[3ex] -2 & 5 & 1 \end{vmatrix} \\[12ex] \underline{\text{3rd Row}} \\[3ex] = -2\begin{vmatrix} 1 & -4 \\[3ex] -15 & -5 \end{vmatrix} -5\begin{vmatrix} 3 & -4 \\[3ex] 6 & -5 \end{vmatrix} + 1\begin{vmatrix} 3 & 1 \\[3ex] 6 & -15 \end{vmatrix} \\[10ex] = -2(-5 - 60) - 5(-15 + 24) + (-45 - 6) \\[3ex] = -2(-65) -5(9) + -51 \\[3ex] = 130 - 45 - 51 \\[3ex] = 34 $
$ \underline{\text{Numerator for }z} \\[3ex] \begin{vmatrix} 3 & 5 & 1 \\[3ex] 6 & -3 & -15 \\[3ex] -2 & 2 & 5 \end{vmatrix} \\[12ex] \underline{\text{2nd Row}} \\[3ex] = -6\begin{vmatrix} 5 & 1 \\[3ex] 2 & 5 \end{vmatrix} + -3\begin{vmatrix} 3 & 1 \\[3ex] -2 & 5 \end{vmatrix} - -15\begin{vmatrix} 3 & 5 \\[3ex] -2 & 2 \end{vmatrix} \\[10ex] = -6(25 - 2) - 3(15 + 2) + 15(6 + 10) \\[3ex] = -6(23) - 3(17) + 15(16) \\[3ex] = -138 - 51 + 240 \\[3ex] = 51 $
$ x = \dfrac{17}{17} = 1 \\[5ex] y = \dfrac{34}{17} = 2 \\[5ex] z = \dfrac{51}{17} = 3 \\[5ex] $ Check
| LHS | RHS |
|---|---|
| $ 3x + 5y - 4z \\[3ex] 3(1) + 5(2) - 4(3) \\[3ex] 3 + 10 - 12 \\[3ex] 1 $ | 1 |
| $ 6x - 3y - 5z \\[3ex] 6(1) - 3(2) - 5(3) \\[3ex] 6 - 6 - 15 \\[3ex] -15 $ | −15 |
| $ -2x + 2y + z \\[3ex] -2(1) + 2(2) + 3 \\[3ex] -2 + 4 + 3 \\[3ex] 5 $ | 5 |
$ Let: \\[3ex] \text{Numerator for}\;x = Matrix\;A \\[3ex] \text{Numerator for}\;y = Matrix\;B \\[3ex] \text{Numerator for}\;z = Matrix\;C \\[3ex] \text{Denominator} = Matrix\;D \\[3ex] $
$ P: (x, y) \rightarrow (-4x - y, 2x) \\[3ex] Q: (x, y) \rightarrow (y, 6x - 9y) \\[3ex] R: (x, y) \rightarrow (x - 2y, 3x + 5y) \\[3ex] $ (a.) Write down the matrices of P, Q and R.
(b.) Find:
(i.) 2P − 3R + Q;
(ii.) QR;
(iii.) the inverse of the matrix R.
$ (a.) \\[3ex] P: (x, y) \rightarrow (-4x - y, 2x) \rightarrow (-4x - y, 2x + 0y) \\[3ex] Q: (x, y) \rightarrow (y, 6x - 9y) \rightarrow (0x + y, 6x - 9y) \\[3ex] R: (x, y) \rightarrow (x - 2y, 3x + 5y) \\[5ex] \underline{\text{Corresponding Matrices}} \\[3ex] P = \begin{bmatrix} -4 & -1 \\[3ex] 2 & 0 \end{bmatrix} \hspace{3em} Q = \begin{bmatrix} 0 & 1 \\[3ex] 6 & -9 \end{bmatrix} \hspace{3em} R = \begin{bmatrix} 1 & -2 \\[3ex] 3 & 5 \end{bmatrix} $
$ (b.) (i.) \\[3ex] 2P = \begin{bmatrix} 2(-4) & 2(-1) \\[3ex] 2(2) & 2(0) \end{bmatrix} = \begin{bmatrix} -8 & -2 \\[3ex] 4 & 0 \end{bmatrix} \\[7ex] 3R = \begin{bmatrix} 3(1) & 3(-2) \\[3ex] 3(3) & 3(5) \end{bmatrix} = \begin{bmatrix} 3 & -6 \\[3ex] 9 & 15 \end{bmatrix} \\[7ex] 2P - 3R + Q = \begin{bmatrix} -8 - 3 + 0 & -2 - (-6) + 1 \\[3ex] 4 - 9 + 6 & 0 - 15 + (-9) \end{bmatrix} = \begin{bmatrix} -11 & 5 \\[3ex] 1 & -24 \end{bmatrix} \\[10ex] (ii.) \\[3ex] QR = \begin{bmatrix} 0 & 1 \\[3ex] 6 & -9 \end{bmatrix}\begin{bmatrix} 1 & -2 \\[3ex] 3 & 5 \end{bmatrix} \\[7ex] = \begin{bmatrix} 0(1) + 1(3) & 0(-2) + 1(5) \\[3ex] 6(1) + -9(3) & 6(-2) + -9(5) \end{bmatrix} \\[7ex] = \begin{bmatrix} 0 + 3 & 0 + 5 \\[3ex] 6 - 27 & -12 - 45 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 & 5 \\[3ex] -21 & -57 \end{bmatrix} $
(ii.) To find the inverse of matrix R, we can use at least two methods.
Use any method you prefer.
$ \underline{\text{1st Approach: By Formula}} \\[3ex] R = \begin{bmatrix} 1 & -2 \\[3ex] 3 & 5 \end{bmatrix} \\[8ex] adj\;R = \begin{bmatrix} 5 & -(-2) \\[2ex] -3 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\[2ex] -3 & 1 \end{bmatrix} \\[8ex] det\;R = \begin{vmatrix} 1 & -2 \\[3ex] 3 & 5 \end{vmatrix} \\[8ex] = 1(5) - 3(-2) \\[3ex] = 5 + 6 \\[3ex] = 11 \\[5ex] R^{-1} = \dfrac{adj\;R}{det\;R} \\[5ex] = \dfrac{\begin{bmatrix} 5 & 2 \\[2ex] -3 & 1 \end{bmatrix}}{11} \\[7ex] = \begin{bmatrix} \dfrac{5}{11} & \dfrac{2}{11} \\[5ex] -\dfrac{3}{11} & \dfrac{1}{11} \end{bmatrix} \\[12ex] \underline{\text{2nd Approach: Row Reduction Method}} \\[3ex] R \ \ | \ \ I = I \ \ | \ \ R^{-1} \\[3ex] \left[ \begin{array}{cc|cc} 1 & -2 & 1 & 0 \\[3ex] 3 & 5 & 0 & 1 \end{array} \right] \underrightarrow{-3R_1 + R_2} \left[ \begin{array}{cc|cc} 1 & -2 & 1 & 0 \\[3ex] 0 & 11 & -3 & 1 \end{array} \right] \\[10ex] \underrightarrow{2R_2 + 11R_1} \left[ \begin{array}{cc|cc} 11 & 0 & 5 & 2 \\[3ex] 0 & 11 & -3 & 1 \end{array} \right] \\[10ex] \underrightarrow{\dfrac{R_1, R_2}{11}} \left[ \begin{array}{cc|cc} 1 & 0 & \dfrac{5}{11} & \dfrac{2}{11} \\[5ex] 0 & 1 & -\dfrac{3}{11} & \dfrac{1}{11} \end{array} \right] \\[10ex] \implies \\[3ex] R^{-1} = \begin{bmatrix} \dfrac{5}{11} & \dfrac{2}{11} \\[5ex] -\dfrac{3}{11} & \dfrac{1}{11} \end{bmatrix} $
Let Matrices P, Q, R be Matrices A, B, C respectively.
