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Solution hints for Exercise 1.1

Ques. No. 9 2 2 1 A = 1/3 -2 1 2 1 -2 2

Before finding determinant , multiply the matrix A by 1/3 , we get

A= 2/3 2/3 1/3 -2/3 1/3 2/3 ------(1) 1/3 -2/3 2/3

Then | A | = 1

For finding determinant value , take 1/3 common from each row in (1) , we get

Cofactor matrix 6 6 3 Aij = 1/9 -6 3 6 3 -6 6

6 -6 3 Adj(A) = 1/9 6 3 -6 3 6 6

A-1 = Adj(A) / |A|

6 -6 3 Therefore A-1 = 1/9 6 3 -6 3 6 6

Taking 3 common from the adjoint matrix, we get

6 -6 3 A-1 = 1/3 6 3 -6 = A 3 6 6

Hence proved. Matrix Inversion Method

Step 1 : Write the given equations in the matrix form AX = B

Step 2 : The solution for matrix inversion method is X = A-1 B

Step 3 : Find the |A| [ solution exists if |A|  0 ]

Step 4 : Find the cofactor matrix and Adjoint

Step 5 : Find A-1 using the formula

A-1 = 1/ |A| . Adj(A)

Step 6 : Multiply A-1 and B and find the values of x , y and z

Determining CONSISTENCY of equations with 3 unknowns using determinant method

If the equations are

a1x + b1y + c1 z = d1 a2x + b2y + c2 z = d2 a3x + b3y + c3 z = d3

then

a1 b1 c1  = a2 b2 c2 a3 b3 c3

d1 b1 c1 x = d2 b2 c2 replacing ‘ x column’ by constants d3 b3 c3

a1 d1 c1 y = a2 d2 c2 replacing ‘ y column’ by constants a3 d3 c3

a1 b1 d1 z = a2 b2 d2 replacing ‘ z column’ by constants a3 b3 d3 TEST FOR CONSISTENCY

Cases Condition Result 1   0 The system of equations is consistent and has unique solution

To solve:

Use cramer’s rule x = x/ ; y = y/ and z = z/ 2 (i)  = 0 The system of equations is consistent (ii) x = y= z = 0 and has infinitely many solutions (iii) atleast one of the 2x2 minors of   0 To solve :

Reduced the given equations to 2 equations and solve using cramer’s rule and getting solutions in terms of ‘k’

3 (i)  = 0 The system of equations is consistent (ii) x = y = z = 0 and has infinitely many solutions (iii) All 2x2 minors of ,x, y and z are zero To solve (iv) but atleast one element in   0 Reduce to one equation and get solution in terms of ‘s’ and ‘t’ 4 (i)  = 0 The system of equations is inconsistent (ii) x or y or z  0 and has NO solution 5 (i)  =0 The system of equations is inconsistent (ii) x = y = z = 0 and has NO solution (iii) All the 2x2 minors of  = 0 (iv) but atleast one 2x2 minor in x or y or z  0

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