Question

To solve the following three equations-
x+2y+z = 7
x + 3z = 11
2x – 3y = 1

Answer 1

Hint: To solve this problem, we should know the basics of solving algebraic equations. We use the technique of matrices to solve this problem. We express the equations in the form of AX = B and then solve for X by taking the inverse of A and then performing the product of the two matrices ( and B) to get the values of x, y and z.

Complete step-by-step answer:

Before solving this problem, we first understand the basics of matrices to solve algebraic equations. We should first know that the number of variables or unknowns (in this case, x, y and z) should be the same as the number of equations available. Clearly, in this case, we have three variables and three equations, thus we can solve these equations to find the values of x, y and z.
In this case, we first express the equations in the form of AX = B. We have,
A = $\left[ \begin{align}
  & {{\text{a}}_{11}}\text{ }{{\text{a}}_{12}}\text{ }{{\text{a}}_{13}} \\
 & {{\text{a}}_{21}}\text{ }{{\text{a}}_{22}}\text{ }{{\text{a}}_{23}} \\
 & {{\text{a}}_{31}}\text{ }{{\text{a}}_{32}}\text{ }{{\text{a}}_{33}} \\
\end{align} \right]$ = $\left[ \begin{align}
  & 1\text{ 2 1} \\
 & \text{1 0 3} \\
 & \text{2 -3 0} \\
\end{align} \right]$ --(1)
We can now get the values of the matrix co-efficient of matrix A by comparing the terms in equation (1).
X = $\left[ \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right]$
B = $\left[ \begin{align}
  & 7 \\
 & 11 \\
 & 1 \\
\end{align} \right]$
Now, X = ${{A}^{-1}}$ B -- (A)
Now, finding ${{A}^{-1}}$, we have,
${{A}^{-1}}$ = $\dfrac{adj(A)}{|A|}$ -- (B)
(where, adj(A) is adjoint of matrix A and |A| is the determinant of A)
If |A| = ${{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}})-{{a}_{12}}({{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}})+{{a}_{13}}({{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}})$ [for 3 $\times $ 3 matrix] -- (2)
For, A = $\left[ \begin{align}
  & {{\text{a}}_{11}}\text{ }{{\text{a}}_{12}}\text{ } \\
 & {{\text{a}}_{21}}\text{ }{{\text{a}}_{22}} \\
\end{align} \right]$, |A| = ${{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$ [for 2 $\times $ 2 matrix]
Now, the formula of adjoint of A is -
adj(A) =$\left[ {\begin{array}{*{20}{c}}
  { + \left| {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right|} \\
  { - \left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{33}}}
\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{13}}} \\
  {{a_{31}}}&{{a_{33}}}
\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{23}}}
\end{array}} \right|} \\
  { + \left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{22}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right|}&{ - \left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right|}&{ + \left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right|}
\end{array}} \right]$
For the given matrix A, we substitute the values from (1), we get,
adj(A) = $\left[ \begin{align}
  & \text{9 -3 6} \\
 & \text{6 -2 -2} \\
 & \text{-3 7 -2} \\
\end{align} \right]$ -- (3)
From using the formula of (2), we have,
|A| = ${{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}})-{{a}_{12}}({{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}})+{{a}_{13}}({{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}})$
Now, substituting the values from (1), we have,
|A| = 18 -- (4)
Thus, from equation (A) and (B), we have,
X = ${{A}^{-1}}$ B
X = $\dfrac{adj(A)}{|A|}$B
X = $\dfrac{{\left[ {\begin{array}{*{20}{c}}
  9&{ - 3}&6 \\
  6&{ - 2}&{ - 2} \\
  { - 3}&7&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  7 \\
  {11} \\
  1
\end{array}} \right]}}{{18}}$
Now, performing the matrix multiplication, we have,
X = $\dfrac{\left[ \begin{align}
  & 9\times 7-3\times 11+6\times 1 \\
 & 6\times 7-2\times 11+-2\times 1 \\
 & -3\times 7+7\times 11-2\times 1 \\
\end{align} \right]}{18}$
X = $\dfrac{\left[ \begin{align}
  & 36 \\
 & 18 \\
 & 54 \\
\end{align} \right]}{18}$
Dividing the elements by 18, we get

X = $\left[ \begin{align}
  & 2 \\
 & 1 \\
 & 3 \\
\end{align} \right]$ = $\left[ \begin{align}
  & x \\
 & y \\
 & z \\
\end{align} \right]$
Thus, the values are x = 2, y = 1, z = 3.

Note: Another simpler way to solve the problem involves solving the equation manually. That is, we can substitute the value of x (in terms of y and z) from one of the equations and put it in other two equations. Then, we can solve the other two equations (since, now they have only 2 variables and 2 equations).