Given system of equations are

x + 2y - 5z = -9

3x -y + 2z = 5

2x + 3y - z = 3

It's matrix form is AX = B

Where A = \(\begin{vmatrix}1&2&-5\\3&-1&2\\2&3&-1\end{vmatrix} \times \begin{vmatrix}x\\y\\z\end{vmatrix} \&\, B = \begin{vmatrix}-9\\5\\3\end{vmatrix}\) 

We have to find inverse of matrix A

∵ A = IA

= \(\begin{vmatrix}1&2&-5\\3&-1&2\\2&3&-1\end{vmatrix} = \begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}A\) 

Applying R₂ → R₂ -3R₁ 

& R₃ → R₃ -2R₁

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&-1&9\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\-2&0&1\end{vmatrix}A\)

Applying R₃ → R₃ - \(\frac 17\) R₂

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&0&\frac {46}{7}\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\\frac {-11}{7}&\frac {-1}{7}&1\end{vmatrix}A\)

Applying R₃ → \(\frac {R_3}{\frac {46}{7}}\) 

\(\begin{vmatrix}1&2&-5\\0&-7&17\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\-3&1&0\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

Applying R₂ → R₂ - 17 R₃ 

\(\begin{vmatrix}1&2&-5\\0&-7&0\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\\frac {43}{46}&\frac {63}{46}&\frac {-119}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

Applying R₂ → \(\frac {R_2}{-7}\)

\(\begin{vmatrix}1&2&-5\\0&1&0\\0&0&1\end{vmatrix} = \begin{vmatrix}1&0&0\\\frac {-7}{46}&\frac {-9}{46}&\frac {17}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

 Applying R₂ → R₁ - 2R₂ + 5R₃

\(\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix} = \begin{vmatrix}\frac {5}{46}&\frac {13}{46}&\frac {1}{46}\\\frac {-7}{46}&\frac {-9}{46}&\frac {17}{46}\\\frac {-11}{46}&\frac {-1}{46}&\frac {7}{46}\end{vmatrix}A\)

∴ A₁ = \(\frac {1}{46}\begin{vmatrix}5&13&1\\-7&-9&17\\-11&-1&7\end{vmatrix} \)

Now, ∵ AX = B

∴ x = A' B

= \(\frac {1}{46}\begin{vmatrix}5&13&1\\-7&-9&17\\-11&-1&7\end{vmatrix} \begin{vmatrix}-9\\5\\3\end{vmatrix}\)

= \(\frac {1}{46}\begin{vmatrix}5\times-9 +13\times5+1\times3\\-7\times -9+(-9) \times 5+ 17\times 3\\-11\times-9 + (-1) \times 5 + 7\times 3\end{vmatrix} = \frac {1}{46}\begin{vmatrix}-45+65+3\\63-45+51\\99-5+21\end{vmatrix}\)

= \(\frac {1}{46}\begin{vmatrix}23\\69\\115\end{vmatrix} = \begin{vmatrix}23/46\\69/46\\115/46\end{vmatrix} = \begin{vmatrix}1/2\\3/2\\5/2\end{vmatrix} \)

Hence, x = \( \begin{vmatrix}x\\y\\z\end{vmatrix} = \begin{vmatrix}1/2\\3/2\\5/2\end{vmatrix} \)

∴ x = \(\frac 12\), y = \(\frac 32\) & z = \(\frac 52\) 

Hence, given system is consistent & its solution is 

  x = \(\frac 12\), y = \(\frac 32\) & z = \(\frac 52\)