Question

How do you solve $6x + 5y = 8\;{\text{and}}\;3x - y = 7$ using matrices ?

Answer 1

Hint:To solve the given equations using matrices, first of all write the given equations in matrix form and then find the cofactor of each term in the first row. After calculating the cofactor of each term, now equate the terms of the first row divided by their respective cofactors with each other. When equating you will get two variable expressions and one constant, so simply equate both variables with that constant separately to get the required solution.

Complete step by step solution:
In order to solve the given equations $6x + 5y = 8\;{\text{and}}\;3x - y = 7$ with help of matrices, we have to first express the given equations in matrix form and for that we have to simplify the equations as follows
$6x + 5y = 8\;{\text{and}}\;3x - y = 7 \\
\Rightarrow 6x + 5y - 8 = 0\;{\text{and}}\;3x - y - 7 = 0 \\ $
Now expressing it in matrix form, we will get
$\left( {\begin{array}{*{20}{c}}
  x&y&1 \\
  6&5&{ - 8} \\
  3&{ - 1}&{ - 7}
\end{array}} \right)$
Finding cofactors of the terms of the first row,
Cofactor of $x = 5( - 7) - ( - 8)( - 1) = - 43$
Cofactor of $y = ( - 8)3 - 6( - 7) = 18$
Cofactor of $1 = 6( - 1) - 3 \times 5 = - 21$
Now, we know that solution of equation in matrix is given as equating the terms of the row divided with their respective cofactors, that is
$ \Rightarrow \dfrac{x}{{ - 43}} = \dfrac{y}{{18}} = \dfrac{1}{{ - 21}}$
Now, equating variable expressions with constant separately, we will get
$\Rightarrow \dfrac{x}{{ - 43}} = \dfrac{1}{{ - 21}}\;{\text{and}}\;\dfrac{y}{{18}}= \dfrac{1}{{ - 21}} \\
\therefore\ x = \dfrac{{43}}{{21}}\;{\text{and}}\;y = - \dfrac{{18}}{{21}} = - \dfrac{6}{7} \\ $
Therefore $x = \dfrac{{43}}{{21}}\;{\text{and}}\;y = - \dfrac{6}{7}$ is the require solution for the given equation.

Note: Take care of the signs (positive and negative) when calculating the cofactors of the terms because their sign also depends on their places that is the sum of their row number and column number, if even then positive and if odd then negative. Also check your by putting it in both the equations, does it satisfy it or not.