Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Matrices & determinants(2*2, 3*3)
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Matrices & determinants(2*2, 3*3)
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Subject: Further Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 1
Theme: Pure Mathematics
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Under stand the array as linear transformation Find determinants and apply to solve simultaneous equations in two or three unknowns
Form the coefficient matrix $D$: $D = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}$ Calculate $|D|$. If $|D|=0$, there is no unique solution (either no solution or infinitely many solutions).
2. Form $D_x$: Replace the first column of $D$ with the constant terms. $D_x = \begin{pmatrix} c_1 & b_1 \\ c_2 & b_2 \end{pmatrix}$ Calculate $|D_x|$.
3. Form $D_y$: Replace the second column of $D$ with the constant terms. $D_y = \begin{pmatrix} a_1 & c_1 \\ a_2 & c_2 \end{pmatrix}$ Calculate $|D_y|$.
4. Solutions: $x = \frac{|D_x|}{|D|}$ $y = \frac{|D_y|}{|D|}$ Worked Example 5: Solving 2x2 Simultaneous Equations (Cramer's Rule) A Nigerian farmer sells yam and plantain. On Monday, a customer bought 3 tubers of yam and 2 bunches of plantain for NGN 4,
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0. On Tuesday, another customer bought 2 tubers of yam and 4 bunches of plantain for NGN 6,
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0. Find the price of one tuber of yam and one bunch of plantain. Let $x$ be the price of one tuber of yam and $y$ be the price of one bunch of plantain.
The equations are: $3x + 2y = 4500$ $2x + 4y = 6000$ Solution:
1. Coefficient matrix $D$: $D = \begin{pmatrix} 3 & 2 \\ 2 & 4 \end{pmatrix}$ $|D| = (3 \times 4) - (2 \times 2) = 12 - 4 = 8$ 2. $D_x$: $D_x = \begin{pmatrix} 4500 & 2 \\ 6000 & 4 \end{pmatrix}$ $|D_x| = (4500 \times 4) - (2 \times 6000) = 18000 - 12000 = 6000$ 3. $D_y$: $D_y = \begin{pmatrix} 3 & 4500 \\ 2 & 6000 \end{pmatrix}$ $|D_y| = (3 \times 6000) - (4500 \times 2) = 18000 - 9000 = 9000$
4. Solutions: $x = \frac{|D_x|}{|D|} = \frac{6000}{8} = 750$ $y = \frac{|D_y|}{|D|} = \frac{9000}{8} = 1125$ Therefore, one tuber of yam costs NGN 750 and one bunch of plantain costs NGN 1,
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5. For a 3x3 System: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$
1. Form the coefficient matrix $D$: $D = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$ Calculate $|D|$. If $|D|=0$, there is no unique solution.
2. Form $D_x$: Replace the first column of $D$ with the constant terms. $D_x = \begin{pmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{pmatrix}$ Calculate $|D_x|$.
3. Form $D_y$: Replace the second column of $D$ with the constant terms. $D_y = \begin{pmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{pmatrix}$ Calculate $|D_y|$.
4. Form $D_z$: Replace the third column of $D$ with the constant terms. $D_z = \begin{pmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{pmatrix}$ Calculate $|D_z|$.
5. Solutions: $x = \frac{|D_x|}{|D|}$ $y = \frac{|D_y|}{|D|}$ $z = \frac{|D_z|}{|D|}$ Worked Example 6: Solving 3x3 Simultaneous Equations (Cramer's Rule) Three agricultural cooperatives (A, B, C) supply bags of maize, beans, and millet to a food processing company. Cooperative A supplies 10 bags of maize, 5 bags of beans, and 3 bags of millet, totaling NGN 150,
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0. Cooperative B supplies 8 bags of maize, 7 bags of beans, and 4 bags of millet, totaling NGN 165,000. * Cooperative C supplies 5 bags of maize, 6 bags of beans, and 2 bags of millet, totaling NGN 110,
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0. Find the cost per bag of maize, beans, and millet. Let $x$ be the cost per bag of maize, $y$ be the cost per bag of beans, and $z$ be the cost per bag of millet.
The equations are: $10x + 5y + 3z = 150000$ $8x + 7y + 4z = 165000$ $5x + 6y + 2z = 110000$ Solution:
1. Coefficient matrix $D$: $D = \begin{pmatrix} 10 & 5 & 3 \\ 8 & 7 & 4 \\ 5 & 6 & 2 \end{pmatrix}$ $|D|$ (using cofactor expansion along first row): $|D| = 10\begin{vmatrix} 7 & 4 \\ 6 & 2 \end{vmatrix} - 5\begin{vmatrix} 8 & 4 \\ 5 & 2 \end{vmatrix} + 3\begin{vmatrix} 8 A. Introduction to Matrices A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. It is typically enclosed in square brackets `[ ]` or parentheses `( )`.
Order (Dimension) of a Matrix: An m x n matrix has m rows and n columns.
Example: A matrix with 2 rows and 3 columns is a 2x3 matrix.
Elements of a Matrix: Each entry in a matrix is called an element. An element `a_ij` refers to the element in the i-th row and j-th column. B. Matrix as a Linear Transformation (for a 2x2 matrix) A 2x2 matrix can represent a linear transformation that maps points from one position to another in a 2D plane. A point `(x, y)` can be written as a column vector `[x; y]`. When this vector is multiplied by a 2x2 transformation matrix, it produces a new point `(x', y')`. Let the transformation matrix `T` be: $T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ And a point `P` be: $P = \begin{pmatrix} x \\ y \end{pmatrix}$ The transformed point `P'` is given by: $P' = T P = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$ Common types of linear transformations represented by 2x2 matrices include: Scaling: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ (scales by factor k)
Rotation: $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ (rotates by angle $\theta$ counter-clockwise)
Reflection: e.g., $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ (reflection in y-axis)
Shear: e.g., $\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ (horizontal shear)
Worked Example 1: Linear Transformation Consider a point $P(2, 1)$ in a 2D plane. Apply the transformation represented by the matrix $M = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$. Find the new coordinates of the point.
Solution: The point P can be written as a column vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. The transformed point $P'$ is calculated as: $P' = M P = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ $P' = \begin{pmatrix} (3 \times 2) + (1 \times 1) \\ (0 \times 2) + (2 \times 1) \end{pmatrix}$ $P' = \begin{pmatrix} 6 + 1 \\ 0 + 2 \end{pmatrix}$ $P' = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$ So, the new coordinates of the point are $(7, 2)$. C. Determinant of a 2x2 Matrix For a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant, denoted as $det(A)$ or $|A|$, is calculated as: $|A| = ad - bc$ Worked Example 2: Determinant of a 2x2 Matrix Calculate the determinant of the matrix $P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$.
Solution: Using the formula $ad - bc$: $|P| = (5 \times 4) - (2 \times 3)$ $|P| = 20 - 6$ $|P| = 14$ D. Determinant of a 3x3 Matrix For a 3x3 matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant can be calculated using two common methods: Method 1: Cofactor Expansion This method involves expanding along a row or a column. The general formula for expanding along the first row is: $|A| = a(e i - f h) - b(d i - f g) + c(d h - e g)$ Minors ($M_{ij}$): The minor of an element $a_{ij}$ is the determinant of the sub-matrix formed by deleting the $i$-th row and $j$-th column. * Cofactors ($C_{ij}$): The cofactor of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.
The signs follow a checkerboard pattern: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ To find the determinant, choose any row or column, multiply each element by its cofactor, and sum the results.
Worked Example 3: Determinant of a 3x3 Matrix (Cofactor Expansion) Calculate the determinant of the matrix $B = \begin{pmatrix} 1 & 2 & 3 \\ deleting the $i$-th row and $j$-th column. * Cofactors ($C_{ij}$): The cofactor of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.
The signs follow a checkerboard pattern: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ To find the determinant, choose any row or column, multiply each element by its cofactor, and sum the results.
Worked Example 3: Determinant of a 3x3 Matrix (Cofactor Expansion) Calculate the determinant of the matrix $B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$.
Solution (Expanding along the first row):
1. Element $a_{11}=1$: Minor $M_{11} = \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = (5 \times 9) - (6 \times 8) = 45 - 48 = -3$ Cofactor $C_{11} = (-1)^{1+1} M_{11} = (+1)(-3) = -3$
2. Element $a_{12}=2$: Minor $M_{12} = \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} = (4 \times 9) - (6 \times 7) = 36 - 42 = -6$ Cofactor $C_{12} = (-1)^{1+2} M_{12} = (-1)(-6) = 6$
3. Element $a_{13}=3$: Minor $M_{13} = \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} = (4 \times 8) - (5 \times 7) = 32 - 35 = -3$ Cofactor $C_{13} = (-1)^{1+3} M_{13} = (+1)(-3) = -3$ Now, sum the products of elements and their cofactors: $|B| = (1 \times C_{11}) + (2 \times C_{12}) + (3 \times C_{13})$ $|B| = (1 \times -3) + (2 \times 6) + (3 \times -3)$ $|B| = -3 + 12 - 9$ $|B| = 0$ Method 2: Sarrus' Rule (Only for 3x3 Matrices)
1. Rewrite the first two columns to the right of the determinant.
2. Multiply along the main diagonals and add the products.
3. Multiply along the reverse diagonals and subtract the products. For matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, Extend: $\begin{array}{ccc|cc} a & b & c & a & b \\ d & e & f & d & e \\ g & h & i & g & h \end{array}$ $|A| = (aei + bfg + cdh) - (ceg + afh + bdi)$ Worked Example 4: Determinant of a 3x3 Matrix (Sarrus' Rule) Calculate the determinant of the matrix $B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$ using Sarrus' Rule.
Solution: Extend the matrix: $\begin{array}{ccc|cc} 1 & 2 & 3 & 1 & 2 \\ 4 & 5 & 6 & 4 & 5 \\ 7 & 8 & 9 & 7 & 8 \end{array}$ Products along main diagonals (top-left to bottom-right): $(1 \times 5 \times 9) = 45$ $(2 \times 6 \times 7) = 84$ $(3 \times 4 \times 8) = 96$ Sum of main diagonal products = $45 + 84 + 96 = 225$ Products along reverse diagonals (top-right to bottom-left): $(3 \times 5 \times 7) = 105$ $(1 \times 6 \times 8) = 48$ $(2 \times 4 \times 9) = 72$ Sum of reverse diagonal products = $105 + 48 + 72 = 225$ $|B| = \text{Sum of main diagonal products} - \text{Sum of reverse diagonal products}$ $|B| = 225 - 225 = 0$ (
Note: Both methods yield the same result, confirming the calculation.)
E. Application: Solving Simultaneous Equations using Cramer's Rule Cramer's Rule is a formula that uses determinants to solve systems of linear equations.
For a 2x2 System: $a_1x + b_1y = c_1$ $a_2x + b_2y = c_2$
1. Form the coefficient matrix $D$: $D = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}$ Calculate $|D|$. If $|D|=0$, there is no unique solution (either no solution or infinitely many solutions).
2. Form $D_x$: Replace the first column of $D$ with the constant terms. $D_x = \begin{pmatrix} c_1 & b_1 \\ c_2 & b_2 \end{pmatrix}$ Calculate $|D_x|$.
3. Form $D_y$: Replace the second column of $D$ with the constant terms. $D_y = \begin{pmatrix} a_1 & c_1 \\ a_2 & c_2 \end{pmatrix}$ Calculate $|D_y|$.
4. Solutions: $x = \frac{|D_x|}{|D|}$ $y = of beans, and $z$ be the cost per bag of millet.
The equations are: $10x + 5y + 3z = 150000$ $8x + 7y + 4z = 165000$ $5x + 6y + 2z = 110000$ Solution:
1. Coefficient matrix $D$: $D = \begin{pmatrix} 10 & 5 & 3 \\ 8 & 7 & 4 \\ 5 & 6 & 2 \end{pmatrix}$ $|D|$ (using cofactor expansion along first row): $|D| = 10\begin{vmatrix} 7 & 4 \\ 6 & 2 \end{vmatrix} - 5\begin{vmatrix} 8 & 4 \\ 5 & 2 \end{vmatrix} + 3\begin{vmatrix} 8 & 7 \\ 5 & 6 \end{vmatrix}$ $|D| = 10(14 - 24) - 5(16 - 20) + 3(48 - 35)$ $|D| = 10(-10) - 5(-4) + 3(13)$ $|D| = -100 + 20 + 39 = -41$ 2. $D_x$: $D_x = \begin{pmatrix} 150000 & 5 & 3 \\ 165000 & 7 & 4 \\ 110000 & 6 & 2 \end{pmatrix}$ $|D_x|$ (using cofactor expansion along first column, factoring out 10000): $|D_x| = 150000\begin{vmatrix} 7 & 4 \\ 6 & 2 \end{vmatrix} - 5\begin{vmatrix} 165000 & 4 \\ 110000 & 2 \end{vmatrix} + 3\begin{vmatrix} 165000 & 7 \\ 110000 & 6 \end{vmatrix}$ It's easier to factor out 10000 first from columns 1 and 3 of the constant terms and calculate. Or use the elements directly. Let's calculate using cofactor expansion along the first row: $|D_x| = 150000(14 - 24) - 5(330000 - 440000) + 3(990000 - 770000)$ $|D_x| = 150000(-10) - 5(-110000) + 3(220000)$ $|D_x| = -1500000 + 550000 + 660000$ $|D_x| = -290000$ 3. $D_y$: $D_y = \begin{pmatrix} 10 & 150000 & 3 \\ 8 & 165000 & 4 \\ 5 & 110000 & 2 \end{pmatrix}$ $|D_y|$ (cofactor expansion along first row): $|D_y| = 10(330000 - 440000) - 150000(16 - 20) + 3(880000 - 825000)$ $|D_y| = 10(-110000) - 150000(-4) + 3(55000)$ $|D_y| = -1100000 + 600000 + 165000$ $|D_y| = -335000$ 4. $D_z$: $D_z = \begin{pmatrix} 10 & 5 & 150000 \\ 8 & 7 & 165000 \\ 5 & 6 & 110000 \end{pmatrix}$ $|D_z|$ (cofactor expansion along first row): $|D_z| = 10(770000 - 990000) - 5(880000 - 825000) + 150000(48 - 35)$ $|D_z| = 10(-220000) - 5(55000) + 150000(13)$ $|D_z| = -2200000 - 275000 + 1950000$ $|D_z| = -525000$
5. Solutions: $x = \frac{|D_x|}{|D|} = \frac{-290000}{-41} \approx 7073.17$ $y = \frac{|D_y|}{|D|} = \frac{-335000}{-41} \approx 8170.73$ $z = \frac{|D_z|}{|D|} = \frac{-525000}{-41} \approx 12804.88$ Given the context of prices, these values should ideally be integers or exact. Let's recheck the determinant calculations to ensure accuracy. A common mistake is arithmetic error or sign errors in cofactors. Rechecking $|D|$: $|D| = 10(14 - 24) - 5(16 - 20) + 3(48 - 35) = 10(-10) - 5(-4) + 3(13) = -100 + 20 + 39 = -41$. This is correct. Let's recheck $|D_x|$: $|D_x| = 150000(14 - 24) - 5(165000 \times 2 - 110000 \times 4) + 3(165000 \times 6 - 110000 \times 7)$ $|D_x| = 150000(-10) - 5(330000 - 440000) + 3(990000 - 770000)$ $|D_x| = -1500000 - 5(-110000) + 3(220000)$ $|D_x| = -1500000 + 550000 + 660000 = -290000$. This is correct. Recheck $|D_y|$: $|D_y| = 10(165000 \times 2 - 110000 \times 4) - 150000(8 \times 2 - 5 \times 4) + 3(8 \times 110000 - 5 \times 165000)$ $|D_y| = 10(330000 - 440000) - 150000(16 - 20) + 3(880000 - 825000)$ $|D_y| = 10(-110000) - 150000(-4) + 3(55000)$ $|D_y| = -1100000 + 600000 + 165000 = -335000$. This is correct. Recheck $|D_z|$: $|D_z| = 10(7 \times 110000 - 6 \times 165000) - 5(8 \times 110000 - 5 \times 165000) + 150000(8 \times 6 - 5 \times 7)$ $|D_z| = 10(770000 - 990000) - 5(880000 - 825000) + 150000(48 - 35)$ $|D_z| = 10(-220000) - 5(55000) + 150000(13)$ $|D_z| = -2200000 - 275000 + 1950000 = -525000$. This is correct. The calculations appear correct. The values are not exact integers, which can happen with real-world data or if the problem was not designed for exact integer solutions. It might indicate that the problem context should be adjusted for cleaner numbers if integer solutions are expected. For a real-world scenario, these decimal values would be acceptable approximations. For the purpose of an
Resource Allocation and Planning in Agriculture: Scenario: A Nigerian state government wants to maximize crop yield across different local government areas (LGAs) by distributing specific quantities of three types of fertilizer. They have historical data on how much yield each fertilizer type contributes to maize, rice, and yam per hectare.
Application: This can be modeled as a system of linear equations, where the unknowns are the optimal quantities of each fertilizer type needed to achieve target yields for each crop. Cramer's Rule can be used to solve for these optimal quantities, helping the Ministry of Agriculture make informed decisions about resource distribution and boost food security.
Example: How many bags of NPK, Urea, and Superphosphate fertilizers should be allocated to achieve specific tons of maize, rice, and yam given known yield contributions per bag? Inventory Management and Cost Analysis for Small Businesses: Scenario: A local distributor of provisions in Onitsha market stocks different brands of sugar, milk, and beverages. They want to determine the unit cost of each item from combined purchase invoices for various bundled orders.
Application: If a customer buys a bundle of 2 packets of sugar, 3 tins of milk, and 1 carton of beverage for a total cost, and another customer buys a different bundle with specific quantities and a total cost, these can form a system of linear equations. Using Cramer's Rule, the distributor can accurately determine the individual cost of each item, helping in pricing and inventory valuation.
Example: A bundle of 3 packets of Indomie, 2 sachets of Milo, and 1 sachet of sugar costs NGN
1
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0. Another bundle of 2 Indomie, 4 Milo, and 3 sugar costs NGN
2
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0. And a third bundle... and so on.
Traffic Flow Management in Urban Areas: Scenario: Managing the flow of vehicles through intersections in a busy city like Lagos or Abuja to minimize congestion. The number of vehicles entering and leaving different sections of an intersection can be represented as a system of linear equations.
Application: By assigning variables to the traffic flow in different lanes and sections, a system of equations can be created to model the traffic dynamics. Although complex systems might require more advanced matrix methods (Gaussian elimination), simpler cases or components can be analyzed using determinant concepts to determine if a unique flow pattern exists or if certain adjustments to traffic lights would lead to bottlenecks (indicated by a zero determinant). This helps urban planners design efficient traffic light timings and road layouts. ---