Question
Question: How do you solve \(5x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11\) using matrices?...
How do you solve 5x+3y=−5and7x+5y=−11 using matrices?
Solution
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−1B. Find the required solution from that equation.
Inverse of a 2×2 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=−5and7x+5y=−11 using matrices, we will first express them in matrix form as
AX=B
Where A,XandB 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 2×2 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×A which will be equal to I2
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=2andy=−5
That is the required set of solutions for the given equations.
Note: The term I2 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
⇒a1ax=a1b⇒x=ab