 Hello and welcome to the session. In this session we are going to discuss the following question which says that Test for consistency and solve the given system of equations x plus 2y plus z is equal to 2, 3x plus 6y plus 5z is equal to 4, 2x plus 4y plus 3z is equal to 3 If we are given three simultaneous equations in three variables x, y and z that is a1x plus b1y plus c1z is equal to d1, a2x plus b2y plus c2z is equal to d2, a3x plus b3y plus c3z is equal to d3 When these system of equations can be written in the matrix form as ax is equal to b where a is the 3 by 3 matrix containing elements a1, a2, a3, b1, b2, b3, c1, c2, c3 x is the 3 by 1 matrix containing elements x, y, z and b is the 3 by 1 matrix containing elements d1, d2, d3 Then if a joint of a into b is equal to 0 then the system of equations is consistent and has infinite number of solutions With this key idea we shall proceed with the solution We are given the following system of equations that is x plus 2y plus z is equal to 2, 3x plus 6y plus 5z is equal to 4, 2x plus 4y plus 3z is equal to 3 The given system can be written in the matrix form as is equal to b where a is the 3 by 3 matrix containing elements 1, 3, 2, 2, 6, 4, 1, 5, 3 b is the 3 by 1 matrix containing elements x, y, z and b is the 3 by 1 matrix containing elements 2, 4, 3 Now determinant of a is given by 1 into 6 into 3 that is 18 minus of 4 into 5 that is 20 minus 2 into 3 into 3 that is 9 minus of 2 into 5 that is 10 plus 1 into 3 into 4 that is 12 minus of 2 into 6 that is 12 which is equal to 1 into 18 minus 20 that is minus 2 minus 2 into 9 minus 10 that is minus 1 plus 1 into 12 minus 12 that is 0 which is equal to 1 into minus 2 that is minus 2 minus 2 into minus 1 that is plus 2 plus 1 into 0 that is 0 which is equal to 0 Now the determinant of a is equal to 0 therefore a inverse does not exist and therefore the system has either no solution and infinite number of solutions Now we shall find a determinant of a which is given by the 3 by 3 matrix containing elements a11, a12, a13, a21, a22, a23, a31, a32, a33 Now a11 is the co-factor of a11 that is 1 which is given by minus 1 raise to the power 1 plus 1 into determinant containing elements 6, 4, 5, 3 which is equal to 1 into 6 into 3 that is 18 minus of 4 into 5 that is 10 minus 20 which is equal to 1 into minus 2 that is minus 2 now a12 is the co-factor of a12 that is 2 and is given by minus 1 raise to the power 1 plus 2 into determinant containing elements 3, 2, 5, 3 which is given by minus 1 raise to the power 3 is minus 1 into 3 into 3 that is 9 minus of 2 into 5 that is 10 which is equal to minus 1 into 9 minus 10 that is minus 1 which is equal to 1 Now a13 is the co-factor of a13 that is 1 and is given by minus 1 raise to the power 1 plus 3 into determinant containing elements 3, 2, 6, 4 which is equal to minus 1 raise to the power 4 that is 1 into 3 into 4 that is 12 minus of 2 into 6 that is 12 which is equal to 1 into 0 that is 0 a21 is the co-factor of a21 that is 3 and is given by minus 1 raise to the power 2 plus 1 into determinant containing elements 2, 4, 1, 3 and is equal to minus 1 raise to power 3 that is minus 1 into 2 into 3 that is 6 minus of 4 into 1 that is 4 which is equal to minus 1 into 6 minus 4 that is 2 which is equal to minus 2 Now a22 is given by the co-factor of a22 that is 6 and is equal to minus 1 raise to the power 2 plus 2 into determinant containing elements 1, 2, 1, 3 and is given by minus 1 raise to power 4 that is 1 into 1 into 3 that is 3 minus of 2 into 1 that is 2 which is equal to 1 into 3 minus 2 that is 1 which is equal to 1 into 1 that is 1 Now a23 is equal to co-factor of a23 that is 5 and is given by minus 1 raise to power 2 plus 3 into determinant containing elements 1, 2, 2, 4 which is given by minus 1 raise to power 5 that is minus 1 into 4 into 1 that is 4 minus of 2 into 2 that is 4 which is 0 and is equal to minus 1 into 0 that is 0 a31 is the co-factor of a31 that is 2 and is equal to minus 1 raise to power 3 plus 1 into determinant containing elements 2, 6, 1, 5 which is equal to minus 1 raise to power 4 that is 1 into 2 into 5 that is 10 minus of 6 into 1 that is 6 which is equal to 1 into 10 minus 6 that is 4 which is given by 4 Now a32 is the co-factor of a32 that is 4 and is given by minus 1 raise to power 3 plus 2 into determinant containing elements 1, 3, 1, 5 and is given by minus 1 raise to power 5 that is minus 1 into 5 into 1 that is 5 minus of 3 into 1 that is 3 and is equal to minus 1 into 5 minus 3 that is 2 which is given by minus 2 a33 is equal to co-factor of a33 that is 3 and is given by minus 1 raise to power 3 plus 3 into determinant containing elements 1, 3, 2, 6 and is given by minus 1 raise to power 6 that is 1 into 1 into 6 that is 6 minus of 3 into 2 that is 6 which is equal to 1 into 6 minus 6 that is 0 1 into 0 is 0 Now we have got the values of a11 that is minus 2, a12 that is 1, a13 that is 0, a21 that is minus 2, a22 that is 1, a23 is 0 a31 is 4, a32 is minus 2, a33 is 0 So a joint of a which is given by the 3 by 3 matrix containing elements a11, a12, a13, a21, a22, a23, a31, a32, a33 can be written as The 3 by 3 matrix containing elements a11 that is minus 2, a12 that is 1, a13 is 0, a21 is minus 2, a22 is 1, a23 is 0, a31 is 4, a32 is minus 2 and a33 is 0 Now we find a joint of a into b Now a joint of a into b is equal to a joint of a is the 3 by 3 matrix containing elements minus 2, 1, 0, minus 2, 1, 0, 4, minus 2, 0 and b is the 3 by 1 matrix containing elements 2 and 4, 3 into b that is the 3 by 1 matrix containing elements 2, 4, 3 Now a joint of a is a 3 by 3 matrix and b is a 3 by 1 matrix So the resultant multiplicant will be a 3 by 1 matrix containing elements minus 2 into 4 plus or minus 2 into 4 plus or 4 into 3, 1 into 2 plus 1 into 4 plus or minus 2 into 3, 0 into 2 plus 0 into 4 plus 0 into 3 is equal to the 3 by 1 matrix containing elements minus 2 into 2 that is minus 4, minus 2 into 4 that is minus 8 plus 4 into 3 that is plus 12 Now 1 into 2 is 2 plus 1 into 4 is 4 minus 2 into 3 is minus 6, 0 into 2 is 0 plus 0 into 4 is 0 plus 0 into 3 is 0 which is equal to the 3 by 1 matrix containing elements minus 12 plus 12, 6 minus 6 and 0 which is equal to the 3 by 1 matrix containing elements 0, 0, 0 that is equal to 0 Therefore a joint of a into b is equal to 0 since determinant of a is 0 and a joint of a into b is 0 The system is consistent infinite number of solutions Now we have the given system of equations as x plus 2y plus z is equal to 2, 3x plus 6y plus 5z is equal to 4, 2x plus 4y plus 3z is equal to 3 If we put x is equal to k and solve any of the two given equations we get 2y plus z is equal to 2 minus k and 6y plus 5z is equal to 4 minus 3k Now if we multiply the first equation by 3 and the second equation by 1 we get 6y plus 3z is equal to 6 minus 3k and the second equation is 6y plus 5z is equal to 4 minus 3k Now subtracting second equation from the first equation we get minus 2z is equal to 2 which implies that z is equal to 2 upon minus 2 which is equal to minus 1 So we get z is equal to minus 1 Now putting the value of z equal to minus 1 in the above equation that is 2y plus z is equal to 2 minus k we get 2y minus 1 is equal to 2 minus k which implies that 2y is equal to 3 minus k that is y is equal to 3 minus k upon 2 Therefore we have got x is equal to k, y is equal to 3 minus k by 2 and z is equal to minus 1 hence the given system has infinite number of solutions Given y is equal to k, y is equal to 3 minus k by 2 and z is equal to minus 1 where k is any constant and this is the required answer This completes our session hope you enjoyed this session