# Exercise Set 1.1 – No. 9

9. In each part, determine whether the given 3-tuple is a solution of the linear system

$\large&space;\\&space;2x_1&&space;-4x_2&&space;-x_3&&space;=&space;1&space;\\&space;x_1&&space;-3x_2&&space;+x_3&&space;=&space;1&space;\\&space;3x_1&&space;-5x_2&&space;-3x_3&&space;=&space;1$

$\large&space;{\color{Blue}&space;(a)&space;\begin{pmatrix}&space;3,&space;&&space;1,&space;&&space;1&space;\end{pmatrix}&space;}$

‘For each part put the values of x1, x2 and x3 a
test whether the values equals the all ones vector in RHS or not.  The matrix

$\large&space;{\color{Blue}&space;A=&space;\begin{bmatrix}&space;2&space;&&space;-4&space;&&space;-1&space;\\&space;1&space;&&space;-3&space;&&space;1&space;\\&space;3&space;&&space;-5&space;&&space;-3&space;\end{bmatrix}&space;}$

The vetor in part (a) is x=(3, 1, 1).

Ax = (1, 1, 1). Hence (3, 1, 1) is a solution.

$\large&space;{\color{Blue}&space;(b)&space;\begin{pmatrix}&space;3,&space;&&space;-1,&space;&&space;1&space;\end{pmatrix}&space;}$

Ax = (9, 7, 11) # (1, 1, 1). Hence (b) is not solution.

$\large&space;{\color{Blue}&space;(c)&space;\begin{pmatrix}&space;13,&space;&&space;5,&space;&&space;2&space;\end{pmatrix}&space;}$

Ax = (4, 0, 8) # (1, 1, 1). Hence (c) is not solution.

$\large&space;{\color{Blue}&space;(d)&space;\begin{pmatrix}&space;\frac{13}{2},&space;&&space;\frac{5}{2},&space;&&space;2&space;\end{pmatrix}&space;}$

Ax = (1, 1, 1) = (1, 1, 1). Hence (d) is a solution.

$\large&space;{\color{Blue}&space;(e)&space;\begin{pmatrix}&space;17,&space;&&space;7,&space;&&space;5&space;\end{pmatrix}&space;}$

Ax = (1, 1, 1) = (1, 1, 1). Hence (e) is a solution.