Question

How do you solve \[x + 2y + 3w + 4z = 10,2x + y + w - z = 1,3x + y + 4w + 3z = 22\] and \[ - 2x + 6y + 4w + 20z = 18\] using matrices?

Answer 1

Hint: A matrix is a collection of numbers organised into a predetermined number of rows and columns. The numbers are usually real numbers. In general, matrices with three rows and three columns can contain complex numbers. However, we make the matrix from the given equations. Here, we use the Gaussian elimination method to solve the problem. Here, Gaussian elimination or gauss-jordan elimination technique is used to solve the system of linear equations with three variables by transforming the equation into an augmented matrix then we need to follow some operations in rows to obtain row-echelon form.

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:
 \[x + 2y + 3w + 4z = 10\] -------(1)
 \[2x + y + w - z = 1\] ----------(2)
 \[3x + y + 4w + 3z = 22\] --------(3)
 \[ - 2x + 6y + 4w + 20z = 18\] -----(4)
Now, write the augmented matrix of the system of linear equations
 \[{\rm A} = \left[ {\begin{array}{*{20}{c}}
  1&2&3&4&{|10} \\
  2&1&1&{ - 1}&{|1} \\
  3&1&4&3&{|22} \\
  { - 2}&6&4&{20}&{|18}
\end{array}} \right] \]
To perform a sequence of row operations until the left hand \[4 \times 4\] identity matrix, we get
Here, we use the gaussian elimination method to solve the system of linear equations that is simplified by getting the echelon form.
Make the pivot in the first column and the first row
Now, Eliminate the first column, using row reduced echelon form
  \[{R_4} \to {R_1} + \dfrac{1}{2}{R_4}\]
 \[ \Rightarrow {\rm A} = \left[ {\begin{array}{*{20}{c}}
  1&2&3&4&{|10} \\
  2&1&1&{ - 1}&{|1} \\
  3&1&4&3&{|22} \\
  0&5&5&{14}&{|19}
\end{array}} \right] \]
Now, Eliminate the 1st element of 3nd row as 1 as 0
 \[{R_3} \to 3{R_1} - {R_3}\]
 \[ \Rightarrow {\rm A} = \left[ {\begin{array}{*{20}{c}}
  1&2&3&4&{|10} \\
  2&1&1&{ - 1}&{|1} \\
  0&5&5&9&{|8} \\
  0&5&5&{14}&{|19}
\end{array}} \right] \]
Now, Eliminate the last element of 4th column as 1 as 0
 \[{R_4} \to 5{R_2} - 3{R_3}\]
 \[ \Rightarrow {\rm A} = \left[ {\begin{array}{*{20}{c}}
  1&2&3&4&{|10} \\
  2&1&1&{ - 1}&{|1} \\
  0&5&5&9&{|8} \\
  0&0&0&5&{|11}
\end{array}} \right] \]
Now, make all element of 2nd row
  \[{R_2} \to 2{R_1} - {R_2}\]
 \[{\rm A} = \left[ {\begin{array}{*{20}{c}}
  1&2&3&4&{|10} \\
  0&3&5&9&{|19} \\
  0&5&5&9&{|8} \\
  0&0&0&5&{|11}
\end{array}} \right] \]
Now, Eliminate the last element of second column to get echelon form
 \[{R_3} \to 5{R_2} - 3{R_3}\]
 \[\left( {\begin{array}{*{20}{c}}
  1&2&3&4&{10} \\
  0&3&5&9&{19} \\
  0&0&{ - 10}&{ - 18}&{ - 71} \\
  0&0&0&5&{11}
\end{array}} \right)\]
Finally, we obtain the row echelon form in this case. So, we get the matrix representation to finding generic solution of the variables,
Now using the back substituting method to get the values of variables \[x\] , \[y\] and \[z\] .
Write the three equation:
 \[x + 2y + 3w + 4z = 10\] ------(5)
 \[3y + 5w + 9z = 19\] --------(6)
 \[ - 10w - 18z = - 71\] -------(7)
 \[5z = 11\] --------(8)
Consider
 \[ \Rightarrow \,\,5z = 11\]
Divide both side by 5, then
 \[\therefore \,\,z = \dfrac{{11}}{5}\]
By substitute the \[z\] value in equation(7), we get
 \[
   \Rightarrow - 10w - 18\left( {\dfrac{{11}}{5}} \right) = - 71 \\
   \Rightarrow - 10w = \dfrac{{198}}{5} - 71 \\
   \Rightarrow w = \dfrac{{198 - 355}}{{5 \times 10}} = \dfrac{{57}}{{50}} \\
 \]
Now, substitute the two values into equation(6), then
 \[
  (3y + 5\left( {\dfrac{{57}}{{50}}} \right) + 9\left( {\dfrac{{11}}{5}} \right) = 19 \\
  3y + \dfrac{{57}}{{10}} + \dfrac{{99}}{5} - 19 = 0 \\
 \]
Take LCM on LHS of the above equation
 \[
  3y + \dfrac{{285 + 990 - 950}}{{50}} = 0 \\
  y = - \dfrac{{285 + 40}}{{50 \times 3}} = - \dfrac{{65}}{{30}} \\
  y = - \dfrac{{13}}{6} \\
 \]
Now, substitute the \[y\] and \[z\] value in equation (4), then
 \[x + 2y + 3w + 4z = 10 \Rightarrow x + 2\left( {\dfrac{{ - 13}}{6}} \right) + 3\left( {\dfrac{{57}}{{50}}} \right) + \dfrac{{11}}{5} = 10\]
Expanding the brackets to simplify, then we get
 \[
  x - \dfrac{{13}}{3} + \dfrac{{11}}{5} = 10 - \dfrac{{171}}{{50}} \\
  x = \dfrac{{329}}{{50}} + \dfrac{{97}}{{15}} = \dfrac{{4875}}{{750}} \\
  x = \dfrac{{13}}{3} \\
 \]
Therefore, The generic solution is \[\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  w \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{13}}{3}} \\
  {\dfrac{{13}}{6}} \\
  {\dfrac{{57}}{{50}}} \\
  {\dfrac{{11}}{5}}
\end{array}} \right] \]

Note: Here, we use gaussian elimination method is used to solve the system of linear equation by transforming the equation into an augmented matrix then we needs to follow some operations in rows to obtain row-echelon form to get the generic solution for the three variables from the matrix representation of the augmented matrix.