Question

Solve the system using the Gauss-Jordan method. 2x – 3y – 9z = 12 4x + 5y – 6z = –14 –5x + 3y – 9z = 21 A. (–3, 7, –5) B. C. No solution D.

Answer 1

`(-3,-2,(-4)/(3))`

Explanation:

We are given the equations,

2x - 3y - 9z = 12

4x + 5y - 6z = - 14

-5x + 3y - 9z = 21

Then, the augmented matrix is given by,
`\begin{bmatrix}2&-3&-9&12\\4&5&-6&-14\\-5&3&-9&21\end{bmatrix}`

Now, we will apply some rules to get the row-echlon form of the matrix.

1.
`R_(1)=>\frac {R_(1)}{2}`.

This gives,
`\begin{bmatrix}1&(-3)/(2)&(-9)/(2)&6\\4&5&-6&-14\\-5&3&-9&21\end{bmatrix}`

2.
`R_(2)=>R_(2)-4R_(1)` and
`R_(3)=>R_(3)+5R_(1)`

We get,
`\begin{bmatrix}1&(-3)/(2)&(-9)/(2)&6\\0&11&12&-38\\0&(-9)/(2)&(-63)/(2)&51\end{bmatrix}`

3.
`R_(2)=>\frac {R_(2)}{11}`

So,
`\begin{bmatrix}1&(-3)/(2)&(-9)/(2)&6\\0&1&(12)/(11)&(-38)/(11)\\0&(-9)/(2)&(-63)/(2)&51\end{bmatrix}`

4.
`R_(3)=>R_(3)+(9)/(2)R_(2)` and
`R_(1)=>R_(1)+(3)/(2)R_(2)`

We get,
`\begin{bmatrix}1&0&(-63)/(22)&(9)/(11)\\0&1&(12)/(11)&(-38)/(11)\\0&0&(-585)/(22)&(390)/(11)\end{bmatrix}`

5.
`R_(3)=>(-22)/(585)* R_(3)`

We get,
`\begin{bmatrix}1&0&(-63)/(22)&(9)/(11)\\0&1&(12)/(11)&(-38)/(11)\\0&0&1&(-4)/(3)\end{bmatrix}`

6.
`R_(1)=>R_(1)+(63)/(22)R_(3)` and
`R_(2)=>R_(2)+(-12)/(11)R_(3)`

So,
`\begin{bmatrix}1&0&0&-3\\0&1&0&-2\\0&0&1&(-4)/(3)\end{bmatrix}`

Thus, we get,

x = -3, y = -2 and z =
`(-4)/(3)`.

Hence, the solution is
`(-3,-2,(-4)/(3))`.