Question

How do you solve $5x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11$ using matrices?

Answer 1

Hint: First of all write the given equations in matrix form $AX = B$ then find the inverse of the matrix A and multiply it to both sides of the equation and do matrix multiplication. After multiplication you will get something $X = {A^{ - 1}}B$. Find the required solution from that equation.
Inverse of a $2 \times 2$ matrix is given as
${\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]$

Complete step by step solution:
In order to solve the given equations $5x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11$ using matrices, we will first express them in matrix form as
$AX = B$
Where $A,\;X\;{\text{and}}\;B$ the matrices which consist of coefficients of variables, variables and constants respectively.
Therefore they can be written as
$A = \left[ {\begin{array}{*{20}{c}}
  5&3 \\
  7&5
\end{array}} \right],\;X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\;{\text{and}}\;B = \left[ {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 11}
\end{array}} \right]$
Writing it in equation form, we will get
$
   \Rightarrow AX = B \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  5&3 \\
  7&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \left[ {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 11}
\end{array}} \right] \\
 $
Now, finding inverse of the matrix A,
We know that the inverse of a $2 \times 2$ matrix is given as
${\left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]$
Therefore inverse of A will be given as follows
$
  {\left[ {\begin{array}{*{20}{c}}
  5&3 \\
  7&5
\end{array}} \right]^{ - 1}} = \dfrac{1}{{5 \times 5 - 3 \times 7}}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right] = \dfrac{1}{{25 - 21}}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right] = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right] \\
  \therefore {A^{ - 1}} = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right] \\
 $
Now multiplying both sides of the equation with inverse of A,
$ \Rightarrow \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  5&3 \\
  7&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  5&{ - 3} \\
  { - 7}&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 11}
\end{array}} \right]$
Since at the left hand side, we have ${A^{ - 1}} \times A$ which will be equal to ${I_2}$
So performing matrix multiplication at right hand side we will get
$
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  {5 \times ( - 5) + ( - 3) \times ( - 11)} \\
  { - 7 \times ( - 5) + 5 \times ( - 11)}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  { - 25 + 33} \\
  {35 - 55}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \dfrac{1}{4}\left[ {\begin{array}{*{20}{c}}
  8 \\
  { - 20}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{8}{4}} \\
  {\dfrac{{ - 20}}{4}}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\; = \left[ {\begin{array}{*{20}{c}}
  2 \\
  { - 5}
\end{array}} \right] \\
 $
Therefore from above matrix equation, we can write
$x = 2\;{\text{and}}\;y = - 5$
That is the required set of solutions for the given equations.

Note: The term ${I_2}$ stands for the identity matrix of order two. Identity matrices have the same properties as the number $1$ in multiplication, that is if we multiply it with any number the number remains the same and that is why in the above equations we have written $IX = X$. Also solving through matrix is same as we solve an equation $ax = b$ that is by multiplying both sides with multiplicative inverse of $a$ as follows
$ \Rightarrow \dfrac{1}{a}ax = \dfrac{1}{a}b \Rightarrow x = \dfrac{b}{a}$