Question

solve the following system of linear equations using matrices.\

$$\begin{cases}7x - 2y = 45\\\\-3x + 5y = 18\\end{cases}$$

select the correct choice below and, if necessary, fill in the answer box to complete your choice.\
\
a. there is one solution. the solution of the system is \boxed{}.\
(simplify your answer. type an ordered pair; using integers or fractions.)\
\
b. there are infinitely many solutions. the solution set is \\{(x,y) | 7x - 2y = 45\\}.\
\
c. the solution set is \varnothing.

Explanation:

Step1: Write the system in matrix form

The system \(

$$\begin{cases}7x - 2y = 45\\-3x + 5y = 18\end{cases}$$

\) can be written as \(AX = B\), where \(A=

$$\begin{bmatrix}7&-2\\-3&5\end{bmatrix}$$

\), \(X=

$$\begin{bmatrix}x\\y\end{bmatrix}$$

\), and \(B=

$$\begin{bmatrix}45\\18\end{bmatrix}$$

\).

Step2: Find the determinant of \(A\)

The determinant of \(A\), denoted as \(\det(A)\), is calculated as:

$$ \det(A)=(7)(5)-(-2)(-3)=35 - 6 = 29 $$

Since \(\det(A)
eq0\), the system has a unique solution.

Step3: Find the inverse of \(A\)

The inverse of a \(2\times2\) matrix \(

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

\) is \(\frac{1}{\det(A)}

$$\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

\). So, the inverse of \(A\) is:

$$ A^{-1}=\frac{1}{29}
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$$

Step4: Multiply \(A^{-1}\) with \(B\) to find \(X\)

$$ X = A^{-1}B=\frac{1}{29}
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$$

First, calculate the first element of \(X\) (which is \(x\)):

$$ x=\frac{1}{29}[(5)(45)+(2)(18)]=\frac{1}{29}[225 + 36]=\frac{261}{29}=9 $$

Then, calculate the second element of \(X\) (which is \(y\)):

$$ y=\frac{1}{29}[(3)(45)+(7)(18)]=\frac{1}{29}[135 + 126]=\frac{261}{29}=9 $$

Answer:

A. There is one solution. The solution of the system is \((9, 9)\)