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Question: How do you solve \(5x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11\) using matrices?...

How do you solve 5x+3y=5  and  7x+5y=115x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11 using matrices?

Explanation

Solution

First of all write the given equations in matrix form AX=BAX = 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=A1BX = {A^{ - 1}}B. Find the required solution from that equation.
Inverse of a 2×22 \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  and  7x+5y=115x + 3y = - 5\;{\text{and}}\;7x + 5y = - 11 using matrices, we will first express them in matrix form as
AX=BAX = B
Where A,  X  and  BA,\;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 2×22 \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 A1×A{A^{ - 1}} \times A which will be equal to I2{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  and  y=5x = 2\;{\text{and}}\;y = - 5
That is the required set of solutions for the given equations.

Note: The term I2{I_2} stands for the identity matrix of order two. Identity matrices have the same properties as the number 11 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=XIX = X. Also solving through matrix is same as we solve an equation ax=bax = b that is by multiplying both sides with multiplicative inverse of aa as follows
1aax=1abx=ba\Rightarrow \dfrac{1}{a}ax = \dfrac{1}{a}b \Rightarrow x = \dfrac{b}{a}