This unit covers the meaning and types of matrices, matrix operations (addition, subtraction, scalar multiplication and multiplication), cofactors, the adjoint and inverse of a matrix, the matrix method for solving simultaneous equations, and business applications of matrix products.
Introduction to Matrices
A set of numbers presented in a rectangular form in rows and columns and enclosed in a square (or round) bracket is known as a matrix.
Matrices are denoted by capital letters such as \(A, B, C\), and their elements by small letters such as \(a_{ij}\), where \(i\) is the row and \(j\) is the column of the element.
The order (or size) of a matrix is written as number of rows × number of columns. A matrix with \(m\) rows and \(n\) columns is of order \(m \times n\).
Matrix vs determinant
| Basis | Matrix | Determinant |
|---|---|---|
| Meaning | An arrangement of numbers in rows and columns | A single numerical value obtained from a square array |
| Notation | Square or round brackets \([\ ]\), \((\ )\) | Vertical bars \(|\ |\) |
| Shape | Can be of any order \(m\times n\) | Always square (\(n\times n\)) |
Types of Matrix
- Row matrix: a matrix having only one row, e.g. \(\begin{bmatrix}1&2\end{bmatrix}\).
- Column matrix: a matrix having only one column, e.g. \(\begin{bmatrix}1\\2\end{bmatrix}\).
- Square matrix: a matrix in which the number of rows equals the number of columns, e.g. \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) (2×2) or \(\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}\) (3×3).
- Null (zero) matrix: a matrix in which every element is zero, e.g. \(O=\begin{bmatrix}0&0\\0&0\end{bmatrix}\).
- Identity (unit) matrix: a square matrix in which every element of the leading diagonal is 1 and all other elements are 0. It is denoted by \(I\), e.g. \(I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\).
- Equal matrices: two matrices are equal if they are of the same order and their corresponding elements are equal.
For example, if \(A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\) and \(B=\begin{bmatrix}1&2\\3&4\end{bmatrix}\), then \(A=B\). But \(\begin{bmatrix}1&3\\2&4\end{bmatrix}\neq B\), because the corresponding elements differ.
Transpose of a matrix
The matrix obtained by interchanging the rows and columns of a matrix \(A\) is called the transpose of \(A\), written \(A'\) or \(A^{T}\). If \(A\) is of order \(m\times n\), then \(A^{T}\) is of order \(n\times m\).
Operations on Matrices
Addition and subtraction
Two matrices can be added or subtracted only if they are of the same order. Corresponding elements are added or subtracted.
Scalar multiplication
Multiplying a matrix by a number \(k\) multiplies every element by \(k\): \(k\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ka&kb\\kc&kd\end{bmatrix}\).
Multiplication of matrices
The product \(AB\) is defined only when the number of columns of \(A\) equals the number of rows of \(B\). If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is \(m\times p\). Each element of \(AB\) is obtained by multiplying the elements of a row of \(A\) with the corresponding elements of a column of \(B\) and adding the products.
With \(A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\) (2×2) and \(B=\begin{bmatrix}3&5&7\\4&6&8\end{bmatrix}\) (2×3), the product \(AB\) is of order 2×3, while \(BA\) is not defined.
Example (2076): find \(AB\)
\(A=\begin{bmatrix}1&3&4\\2&1&3\end{bmatrix}\) (2×3) and \(B=\begin{bmatrix}2&4&6\\1&3&5\\2&1&3\end{bmatrix}\) (3×3), so \(AB\) is 2×3.
Example (2074): show that \(AB\neq BA\)
\(A=\begin{bmatrix}4&2\\-1&8\end{bmatrix}\), \(B=\begin{bmatrix}1&3\\5&0\end{bmatrix}\).
Hence \(AB\neq BA\): matrix multiplication is, in general, not commutative.
Example (2075): find \(A^{2}-4A-5I\)
\(A=\begin{bmatrix}1&3\\2&4\end{bmatrix}\).
Cofactor, Adjoint and Inverse
Minor and cofactor
The minor \(M_{ij}\) of an element \(a_{ij}\) is the determinant left after deleting the \(i\)-th row and \(j\)-th column. The cofactor is the signed minor:
The signs follow a chessboard pattern:
Adjoint of a matrix
The adjoint of a square matrix \(A\) (written Adj \(A\)) is the transpose of the matrix of cofactors. For a 2×2 matrix, interchange the leading-diagonal elements and change the signs of the other two:
Example (2072): find Adj \(A\) for \(A=\begin{bmatrix}1&2\\3&-5\end{bmatrix}\)
Cofactors: \(A_{11}=-5\), \(A_{12}=-3\), \(A_{21}=-2\), \(A_{22}=1\).
Singular and non-singular matrices
A square matrix \(A\) is singular if \(|A|=0\) and non-singular if \(|A|\neq0\). Only a non-singular matrix has an inverse.
Inverse of a matrix
If \(AB=BA=I\), then \(B\) is the inverse of \(A\), written \(A^{-1}\). Steps to find \(A^{-1}\):
- Find \(|A|\). If \(|A|\neq0\), \(A^{-1}\) exists.
- Find the cofactor of each element.
- Form the matrix of cofactors.
- Transpose it to get Adj \(A\).
- Divide by the determinant.
Example (2073): inverse of \(A=\begin{bmatrix}3&2\\5&7\end{bmatrix}\)
Example (2072): inverse of \(A=\begin{bmatrix}2&1\\1&6\end{bmatrix}\)
Example (2076): show that \(A\) and \(B\) are inverses of each other
\(A=\begin{bmatrix}3&2\\1&1\end{bmatrix}\), \(B=\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\).
Since \(AB=BA=I\), \(B=A^{-1}\) and \(A=B^{-1}\).
Example (2061): inverse of a 3×3 matrix
\(A=\begin{bmatrix}5&2&1\\2&4&2\\-1&1&3\end{bmatrix}\).
Example (2073): find \(A^{-1}\) and verify \(AA^{-1}=A^{-1}A=I\)
\(A=\begin{bmatrix}5&2&-1\\2&4&1\\1&2&3\end{bmatrix}\).
Similarly, \(A^{-1}A=I\).
Example: a singular matrix
For \(A=\begin{bmatrix}1&1&2\\2&4&6\\3&2&5\end{bmatrix}\):
Since \(|A|=0\), \(A\) is singular and \(A^{-1}\) does not exist.
Matrix Method for Simultaneous Equations
A system of linear equations can be written as \(AX=B\), where \(A\) is the coefficient matrix, \(X\) the column of unknowns and \(B\) the column of constants. If \(|A|\neq0\):
Two variables
Example (2072): solve \(6x+8y=30,\ 9x+5y=42\)
Hence \(x=\tfrac{31}{7}\), \(y=\tfrac{3}{7}\).
Three variables
Example (2074): solve \(x+2y+3z=20,\ 3x+6y+2z=32,\ x+y+z=9\)
Hence \(x=2\), \(y=3\), \(z=4\).
Example (2063): solve \(x+y+z=8,\ x-y+2z=6,\ 3x+5y-7z=14\)
Hence \(x=5\), \(y=\tfrac53\), \(z=\tfrac43\).
Example (2069): production on three machines
Tables (\(x\)), desks (\(y\)) and chairs (\(z\)) need machine hours on machines \(M_1, M_2, M_3\) as given by the equations below (available hours on the right). Find the number of each item produced.
The firm produces 5 tables, 4 desks and 6 chairs.
Example (2066): three products on three machines
Products \(P_1, P_2, P_3\) (quantities \(x, y, z\)) use machine time as follows:
Hence \(x=6\), \(y=5\), \(z=4\) units.
Business Applications of Matrix Products
When quantities are arranged in one matrix and rates (prices, costs, profits) in another, their product gives totals such as revenue, cost or profit. Arrange the matrices so that the columns of the first match the rows of the second.
Raw-material cost of laptops (2074)
Laptop models A, B and C need units of raw materials \(R_1, R_2, R_3\) as below. The cost per unit of \(R_1, R_2, R_3\) is Rs 5000, 6000 and 4000.
| Model | \(R_1\) | \(R_2\) | \(R_3\) |
|---|---|---|---|
| A | 40 | 20 | 10 |
| B | 50 | 15 | 15 |
| C | 30 | 12 | 12 |
Cost per laptop: A Rs 3,60,000; B Rs 4,00,000; C Rs 2,70,000. For an order of 10, 9 and 6 laptops, the total raw-material cost is Rs 88,20,000.
Cost of drugs in different markets (2072)
Matrix \(A\) gives the chemicals \(C_1, C_2, C_3\) (rows) needed per kg of drugs \(D_1, D_2\) (columns). Matrix \(B\) gives the prices of \(C_1, C_2, C_3\) (columns) in three markets (rows).
Each row of \(BA\) gives the cost of \(D_1\) and \(D_2\) per kg in that market.
Monthly profit of a biscuit company (2072)
Two plants produce Tiger, Glucose and Salty biscuits daily (packets). Plant I works 20 days and Plant II works 25 days a month. Cost/selling price per packet: Tiger Rs 10/13, Glucose Rs 11/12, Salty Rs 12/14.
Monthly production is 11,000 Tiger, 11,500 Glucose and 7,000 Salty packets; total monthly profit is Rs 58,500.
Cost of vitamin tablets (2067)
The cost per unit of the three products is 825, 795 and 890; the total cost of the required quantities is Rs 24,405.
Cost per product (2065)
The costs of products A, B and C are Rs 87, Rs 60 and Rs 69.
Stock and sales of cold drinks (2064)
Two shops A and B hold stock \(P\) of three brands and sell quantities \(R\). Prices per bottle are Rs 13, 14 and 15.
\(P-R\) is the stock left. Sales revenue: shop A Rs 598, shop B Rs 930.
Best city for two merchants (2064)
Matrix \(P\) gives the stock of items X, Y, Z (rows) held by merchants A and B (columns). Matrix \(Q\) gives the prices of X, Y, Z (columns) in cities I, II and III (rows).
Both merchants earn the most in City III (Rs 3,73,500 for A and Rs 3,77,000 for B), so both should sell there.
Profit from warehouse sales (2061)
Matrix \(P\) gives the stock of three items (rows) in three warehouses (columns); \(Q\) gives the stock remaining after sales. The profit per unit of the three items is Rs 10, 5 and 15.
Total profit is Rs 29,09,140. After deducting the given expense of Rs 50,000, the net profit is Rs 28,59,140.
Revenue, cost and gross profit in two markets (2059)
Matrix \(A\) gives the sales of products P, Q, R in markets I and II. Selling prices are Rs 2.50, 1.25, 1.50 and cost prices Rs 1.80, 1.20, 0.80 per unit.
Total gross profit from both markets is Rs 27,900.
Total cost of production (2058)
The per-unit costs of products A, B and C are given under three cost heads (raw material, a second cost head and labour). Quantities produced are 300, 250 and 150.
Total cost \(=48750+1075000+510000=\) Rs 16,33,750.
Choosing a production method (2054)
Three methods produce units of three products as shown; the profit per unit of the products is Rs 100, 145 and 80.
Method I gives the highest profit (Rs 3,060), so it should be adopted.
Important Questions
Board exam questions asked on this unit, by exam year (marks in brackets). Short answers are given where available.
2078 Exam
Find \(7(A-B)\) where \(A=\begin{bmatrix}7&8\\9&10\\2&5\end{bmatrix}\) and \(B=\begin{bmatrix}6&3\\3&2\\7&5\end{bmatrix}\). [2 marks]
2079 Exam
Find \(7(A+B)\) where \(A=\begin{bmatrix}1&2&5\\2&1&3\\3&3&2\end{bmatrix}\) and \(B=\begin{bmatrix}3&1&5\\2&5&1\\1&3&2\end{bmatrix}\). [2 marks]
2080 Exam
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Find the transpose of matrix \(A\) where \(A=\begin{bmatrix}1&-2\\4&3\end{bmatrix}\). [2 marks]
$$A^{T}=\begin{bmatrix}1&4\\-2&3\end{bmatrix}$$
2081 Exam
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Find \(\tfrac12(A-B)\) where \(A=\begin{bmatrix}8&9\\9&10\\2&5\end{bmatrix}\) and \(B=\begin{bmatrix}5&6\\3&4\\8&5\end{bmatrix}\). [2 marks]
$$\tfrac12(A-B)=\tfrac12\begin{bmatrix}3&3\\6&6\\-6&0\end{bmatrix}=\begin{bmatrix}\tfrac32&\tfrac32\\3&3\\-3&0\end{bmatrix}$$
2082 Exam
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Find \(3(A+B)\) where \(A=\begin{bmatrix}1&9\\4&1\\3&7\end{bmatrix}\) and \(B=\begin{bmatrix}2&6\\3&3\\1&4\end{bmatrix}\). [2 marks]
$$3(A+B)=3\begin{bmatrix}3&15\\7&4\\4&11\end{bmatrix}=\begin{bmatrix}9&45\\21&12\\12&33\end{bmatrix}$$
2083 Exam
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If \(A=\begin{bmatrix}1&4\\3&2\\2&5\end{bmatrix}\) and \(B=\begin{bmatrix}-1&-2\\0&5\\3&1\end{bmatrix}\), find \(A+B\). [2 marks]
$$A+B=\begin{bmatrix}0&2\\3&7\\5&6\end{bmatrix}$$ -
Solve the equations by the matrix or determinant method: \(x_1+3x_2+4x_3=7,\ 4x_1+2x_2+3x_3=10,\ 2x_1+2x_2+2x_3=6\). [10 marks]
$$\begin{aligned} |A|&=1(4-6)-3(8-6)+4(8-4)=-2-6+16=8\\ \operatorname{Adj}A&=\begin{bmatrix}-2&2&1\\-2&-6&13\\4&4&-10\end{bmatrix}\\ X&=\frac18\begin{bmatrix}-14+20+6\\-14-60+78\\28+40-60\end{bmatrix}=\frac18\begin{bmatrix}12\\4\\8\end{bmatrix} \end{aligned}$$Hence \(x_1=\tfrac32\), \(x_2=\tfrac12\), \(x_3=1\).
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(a) A box contains 4 black and 4 red balls. If 2 balls are drawn at random, find the probability that both are black. [10 marks]
(b) From the given payoff matrix, give the decision according to the minimax regret criterion:
Course of action State of nature (demand) \(S_1\) – 100 500 500 500 \(S_2\) – 100 300 550 550 \(S_3\) – 100 100 350 600
Additional practice
Solve \(3x+5y-4z=-1000,\ -3x+2y+z=0,\ 4x-y+6z=40000\).
Answer: \(x=3000\), \(y=2000\), \(z=5000\). (Check: \(9000+10000-20000=-1000\).)
Solve \(x+y+z=1200,\ 12x+16y+14z=18480,\ z=2x\).
Answer: \(x=90\), \(y=930\), \(z=180\).