Option A:
[tex]\left[\begin{array}{c}x\\y\\\end{array}\right]=\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right] \left[\begin{array}{c}15\\18\\\end{array}\right][/tex]
Solution:
Given system of equations are
[tex]\left\{\begin{array}{r}2 x+5 y=15 \\x+3 y=18\end{array}\right.[/tex]
We can write this equation in matrix form.
Isolate the variables in the equation.
[tex]\left[\begin{array}{cc}2&5\\1&3\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right]=\left[\begin{array}{c}15\\18\\\end{array}\right][/tex]
Where [tex]A=\left[\begin{array}{cc}2&5\\1&3\\\end{array}\right] , X= \left[\begin{array}{c}x\\y\\\end{array}\right] \ \text{and} \ B =\left[\begin{array}{c}15\\18\\\end{array}\right][/tex]
This is in the form of AX = B.
[tex]\Rightarrow X= A^{-1} B[/tex] – – – (1)
Let us first calculate inverse of A.
[tex]$A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\-c & a\end{array}\right][/tex]
[tex]$A^{-1}=\frac{1}{2\times 3-5 \times1}\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right][/tex]
[tex]$A^{-1}=\frac{1}{6-5 }\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right][/tex]
[tex]$A^{-1}=\frac{1}{1 }\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right][/tex]
[tex]$A^{-1}=\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right][/tex]
Substitute this in (2), we get
[tex]\left[\begin{array}{c}x\\y\\\end{array}\right]=\left[\begin{array}{cc}3 & -5 \\-1 & 2 \end{array}\right] \left[\begin{array}{c}15\\18\\\end{array}\right][/tex]
Hence option A is the correct answer.
