When a system of linear equation is consistent and inconsistent? Give an example for each. Test the consistency and solve: x + y + z = 4, x + 2y + 2z = 2, 2x + 2y + z = 5.

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Suresh Chand · Accepted Answer

A system of linear Equation is consistent if the right most column of the augmented matrix is not a pivot column i.e. if Echelon form of the augmented matrix has not row of the augmented matrix has not row of the form [0  0   ...............  0 : k] where k  is non-zero value. otherwise, it is said to be inconsistent Let us consider two augmented matrices in Echelon form (begin{bmatrix}1 & 2 & 3 & : & 2 0 & 3 & 4 & : & 3 0 & 0 & 0 & : & 8end{bmatrix}) It is Inconsistent. (begin{bmatrix}1 & 3 & : & 8 0 & 1 & : & 3 0 & 3 & : & 6end{bmatrix}) It is Consistent. Given system of linear Equation: x + y + z = 4 x + 2y + 2z = 2 2x + 2y + z = 5 writing in matrix form (begin{bmatrix}1 & 1 & 1 1 & 2 & 2 2 & 2 & 1end{bmatrix}) (begin{bmatrix}x y zend{bmatrix}) = (begin{bmatrix}4 2 5end{bmatrix}) The augmented matrix is (begin{bmatrix}1 & 1 & 1 & : & 4 1 & 2 & 2 & : & 2 2 & 2 & 1 & : & 5end{bmatrix}) Apply R2 → R2 - R1,   R3 → R3 - 2R1 ~ (begin{bmatrix}1 & 1 & 1 & : & 4 0 & 1 & 1 & : & -2 0 & 0 & -1 & : & -3end{bmatrix}) clearly, We can see the system is consitent. Now, row reducing it to reduced Echelon form Apply R1 → R1 - R2,   R3 → (-1) R3 ~ (begin{bmatrix}1 & 0 & 0 & : & 6 0 & 1 & 1 & : & -2 0 & 0 & 1 & : & 3end{bmatrix}) Apply R2 → R2 - R3 ~ (begin{bmatrix}1 & 0 & 0 & : & 6 0 & 1 & 0 & : & -5 0 & 0 & 1 & : & 3end{bmatrix}) From this we get x = 6 y = -5 z = 3

When a system of linear equation is consistent and inconsistent? Give an example for each. Test the consistency and solve: x + y + z = 4, x + 2y + 2z = 2, 2x + 2y + z = 5.

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When a system of linear equation is consistent and inconsistent? Give an example for each. Test the consistency and solve: x + y + z = 4, x + 2y + 2z = 2, 2x + 2y + z = 5.

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