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**Hint**: Here in this question, we have to solve the variable \[x\], \[y\], and \[z\] using matrices. The method of solving this type of question is known as “Gauss-Jordan elimination”. First, we have to construct an augmented matrix by using the coefficients of variables and later by the row echelon form and using the back substitution method we get the required solution.

**Complete step by step solution:**

The Gauss-Jordan method, also known as Gauss-Jordan elimination method is used to solve a system of linear equations and is a modified version of Gauss Elimination Method.

we have to perform 2 different process in Gauss Elimination Method i.e.,

1) Formation of upper triangular matrix, and

2) Back substitution

using reduced row echelon form.

Consider the given system of linear equations:

\[2x + 5y - 2z = 14\]--------(1)

\[5x - 6y + 2z = 0\]--------(2)

\[4x - y + 3z = - 7\]--------(3)

Now, write the augmented matrix of the system of linear equations

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

2&5&{ - 2}&|&{14} \\

5&{ - 6}&2&|&0 \\

4&{ - 1}&3&|&{ - 7}

\end{array}} \right]\]

Make the pivot in the first column and the first row

Now, Eliminate the elements in matrix step by step, using row reduced echelon form

\[{R_3} \to {R_3} - 2{R_1}\;\]

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

2&5&{ - 2}&|&{14} \\

5&{ - 6}&2&|&0 \\

0&{ - 11}&7&|&{ - 35}

\end{array}} \right]\]

\[{R_1} \to 5{R_1}\] and \[{R_2} \to 2{R_2}\;\]

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

{10}&{25}&{ - 10}&|&{70} \\

{10}&{ - 12}&4&|&0 \\

0&{ - 11}&7&|&{ - 35}

\end{array}} \right]\]

\[{R_2} \to {R_2} - {R_1}\]

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

{10}&{25}&{ - 10}&|&{70} \\

0&{ - 37}&{14}&|&{ - 70} \\

0&{ - 11}&7&|&{ - 35}

\end{array}} \right]\]

\[{R_2} \to 11{R_2}\] and \[{R_3} \to 37{R_3}\]

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

{10}&{25}&{ - 10}&|&{70} \\

0&{ - 407}&{154}&|&{ - 770} \\

0&{ - 407}&{259}&|&{ - 1295}

\end{array}} \right]\]

\[{R_3} \to {R_3} - {R_2}\]

\[ \Rightarrow \,A = \left[ {\begin{array}{*{20}{c}}

{10}&{25}&{ - 10}&|&{70} \\

0&{ - 407}&{154}&|&{ - 770} \\

0&0&{105}&|&{ - 525}

\end{array}} \right]\]

Now using the back substituting method to get the values of variables \[x\], \[y\] and \[z\].

Write the three equation:

\[10x + 25y - 10z = 70\]-----(4)

\[ - 407y + 154z = - 770\]------(5)

\[105z = - 525\]----------(6)

Let us take equation (6)

\[ \Rightarrow \,\,105z = - 525\]

Divide both side by 105, then

\[ \Rightarrow \,\,z = - \dfrac{{525}}{{105}}\]

\[ \Rightarrow \,\,z = - 5\]

Hence, the value of \[z = - 5\]

Substitute the \[z\] value in equation (5), then

\[ \Rightarrow \,\, - 407y + 154\left( { - 5} \right) = - 770\]

\[ \Rightarrow \,\, - 407y - 770 = - 770\]

Add 770 on both side, then we have

\[ \Rightarrow \,\, - 407y = - 770 + 770\]

\[ \Rightarrow \,\, - 407y = 0\]

Divide both side by -407, the n

\[ \Rightarrow \,\,y = \dfrac{0}{{ - 407}}\]

\[\therefore \,\,y = 0\]

Now, substitute the \[y\] and \[z\] value in equation (4), then

\[ \Rightarrow \,\,10x + 25\left( 0 \right) - 10\left( { - 5} \right) = 70\]

\[ \Rightarrow \,\,10x + 0 + 50 = 70\]

Subtract 50 on both side, then we get

\[ \Rightarrow \,\,10x = 70 - 50\]

\[ \Rightarrow \,\,10x = 20\]

Divide 10 on both side, then

\[ \Rightarrow \,x = \dfrac{{20}}{{10}}\]

\[\therefore \,\,x = 2\]

Hence, the required solution is

\[\left[ {\begin{array}{*{20}{c}}

x \\

y \\

z

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

2 \\

0 \\

{ - 5}

\end{array}} \right]\]

**Note**: When solving this type of questions, when an augmented matrix contains the coefficients of the unknowns and the "pure" coefficients. You can manipulate the rows of this matrix (elementary row operations) to transform the coefficients and to "read", at the end, the solutions of your system. And while solving the back substitution method we take the equation from the bottom of the augmented matrix.

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