Question

How do you solve 3x + 2y + 2z = 3, x + 2y – z = 5 and 2x – 4y + z = 0 using matrices?

Answer 1

Hint: We will first write them in matrix form and after that we will try to convert them in Echelon form and thus use back substitution to find the required values.

Complete step-by-step answer:
Since, we are given that we need to solve the equations 3x + 2y + 2x = 3, x + 2y – z = 5 and 2x – 4y + z = 0 using matrices.
Writing them in matrix form, we will obtain the following equation:-
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {3x + 2y + 2z} \\
  {x + 2y - z} \\
  {2x - 4y + z}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3 \\
  5 \\
  0
\end{array}} \right]\]
Breaking them apart and making an augmented matrix, we can write them as:-
\[ \Rightarrow \left[ {\left. {\begin{array}{*{20}{c}}
  3&2&2 \\
  1&2&{ - 1} \\
  2&{ - 4}&1
\end{array}} \right|\begin{array}{*{20}{c}}
  3 \\
  5 \\
  0
\end{array}} \right]\]
Doing the transformation: ${R_2} \leftrightarrow {R_1}$, we will get:-
\[ \Rightarrow \left[ {\left. {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  3&2&2 \\
  2&{ - 4}&1
\end{array}} \right|\begin{array}{*{20}{c}}
  5 \\
  3 \\
  0
\end{array}} \right]\]
Now, we will use the transformation: ${R_2} \to {R_2} - 3{R_1}$ and ${R_3} \to {R_3} - 2{R_1}$, we will get:-
\[ \Rightarrow \left[ {\left. {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  0&{ - 4}&5 \\
  0&{ - 8}&3
\end{array}} \right|\begin{array}{*{20}{c}}
  5 \\
  { - 12} \\
  { - 10}
\end{array}} \right]\]
Now, we will use the transformation: ${R_3} \to {R_3} - 2{R_2}$, we will get:-
\[ \Rightarrow \left[ {\left. {\begin{array}{*{20}{c}}
  1&2&{ - 1} \\
  0&{ - 4}&5 \\
  0&0&{ - 7}
\end{array}} \right|\begin{array}{*{20}{c}}
  5 \\
  { - 12} \\
  {14}
\end{array}} \right]\]
Now if we see the last row, we see that – 7 z = 14.
If we divide both the equations by – 7, we will then get: z = - 2.
Now, looking at the last row, we have – 4 y + 5 z = - 12
Putting z = - 2, we will get: - 4 y – 10 = - 12
Taking 10 from subtraction in the left hand side to addition in the right hand side, we will then get:-
$ \Rightarrow $- 4 y = - 12 + 10
Simplifying the above equation by clubbing the terms on the right hand side, we will then obtain the following equation:-
$ \Rightarrow $- 4 y = - 2
Dividing both sides by -4, we will then obtain: $y = \dfrac{1}{2}$
Now looking at the first row, we have: x + 2y – z = 5
Putting y and z, we will get: x + 1 – ( - 2) = 0
Simplifying it, we will get: x + 3 = 0
Solving it, we will then obtain: x = - 3

Thus, the answer is x = -3, y = 1/2 and z = -2.

Note:
The students must note that if they do the column transformation, they will also have to make the other transformations because columns do invert the x’s, y’s and z’s. Therefore, we prefer to use row transformations because they do not create any such problems.
The students must also know that we cannot use column transformation and row transformation intertwined. Either you will use the row transformations overall in the whole solution or use the column transformations in the whole.