Linear Algebra
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These are the solutions to the WASSCE Further Mathematics past questions on Linear Algebra.
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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
$
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
$
$ \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] $
