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Quantitative Methods Session 16 – 22.08.13 Matrices Pranjoy Arup Das. Matrices or Matrix A matrix is table of numbers listed in a square or rectangular form. Matrices are used in the following applications: Solving linear equations with 2 variables
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Quantitative MethodsSession 16 – 22.08.13MatricesPranjoy Arup Das
Matrices or Matrix • A matrix is table of numbers listed in a square or rectangular form. • Matrices are used in the following applications: Solving linear equations with 2 variables Solving linear equations with 3 variables Set Theory – To solve problems of Relation of sets Cryptology – the study of coding and de-coding secret messages. Input – output analysis Forecasting related studies Transition probability analysis
This is a 2 x 2 matrix. It has 2 rows and two columns • This is a 2 x 3 matrix. It has 2 rows and 3 columns • This is a 3 x 4 matrix. It has 3 rows and 4 columns. • Matrices are usually denoted by letters such a A, B, C, X etc
Addition of matrices: • Two Matrices having the same number of rows and columns can be added. • Suppose A = • And B = • So A + B = + = = • Rule to remember : A + B = B + A
Multiplication of matrices: • Multiplication is only possible if the rows of the 1st matrix and the column of the 2nd matrix have the same number of elements. • Suppose A = • And B = • So A * B = * = • a = 2 * 5 + 1 * 4 • b = 2 * 6 + 1 * 8 • c = 3 * 5 + 12 * 4 • d = 3 * 6 + 12 * 8 • Rule to remember : A * B is not the same as B * A
Determinant of a matrix: Say A = Then determinant of A or I A I = (ad – bc) Eg. Find the determinant of A = I A I = 5 * 8 – 6 * 4= 40 – 24 = 16 The determinant of = 16
Inverse of a matrix: Say A = Inverse is denoted by 1/A or A-1 NOTE : Inverse of a matrix A, i.e. A-1 = Inverse is not possible if (ad- bc) = 0 Eg. Find the inverse of A = A-1 = = =
Solving simultanoeus linear equations with 2 variables: Suppose we have two equations 2x-y = 3 and 5x+y = 4 We can represent these two equations in the form of 3 matrices: • A, which is the co-efficient matrix = • X, which is the variable matrix = 3) B, which is the constant matrix = The solution is given by AX = B So =
Now Since AX = B => X = 1/A * B => X = A-1 * B We know that A-1 = = = Using the value of A-1 in X = A-1 * B X = => = = > = => = => = So x = 1 and y = -1
Practice Exercise: 1) Try solving the following three equations by matrix method: 2x + y = 15, 2y + z = 25 & 2z+x = 26 2) Find the inverse of the matrix A = [7 5] [6 6] 3) Find the determinant of the matrix: [7 5] [6 6] 4) Find the product of the two matrices [1 3] & [1 2] [2 2] [2 1] 5) Find the sum of the two matrices: [1 2] & [1 3] [2 1] [2 2]