[tex]=\text{ }\begin{pmatrix}-27 \\ 8\end{pmatrix}\text{ (option A)}[/tex]Explanation:[tex]\begin{gathered} \\ \begin{pmatrix}1 & 7 \\ 3 & 4\end{pmatrix}\text{ }\times\text{ x = }\begin{pmatrix}2 \\ 5\end{pmatrix} \end{gathered}[/tex][tex]\begin{gathered} \text{let x =}\begin{pmatrix}x \\ y\end{pmatrix} \\ \begin{pmatrix}2 & 7 \\ 1 & 4\end{pmatrix}\text{ }\times\begin{pmatrix}x \\ y\end{pmatrix}\text{ = }\begin{pmatrix}2 \\ 5\end{pmatrix} \\ mu\text{ltiplying the matrix:} \\ 2(x)\text{ + 7(y) = 2} \\ 2x\text{ + 7y = 2 }\ldots equation1 \\ 1(x)\text{ + 4(y) = 5} \\ x\text{ + 4y = 5 }\ldots equation2 \end{gathered}[/tex]
Using elimination method:
multiply equation 2 by 2
2(x) + 2(4y) = 2(5)
2x + 8y = 10 ...equation 2
combine both equations:
2x + 7y = 2 ...equation 1
2x + 8y = 10 ...equation 2
subtract equation 2 from 1:
2x - 2x + 7y - 8y = 2 - 10
0 - y = -8
-y = -8
y = -8/-1
y = 8
substitute for y in equation 2:
x + 4y = 5
x + 4(8) = 5
x + 32 = 5
x = 5 - 32
x = -27
[tex]\begin{gathered} \text{Thesolution of the matrix = }x\text{ =}\begin{pmatrix}x \\ y\end{pmatrix} \\ =\text{ }\begin{pmatrix}-27 \\ 8\end{pmatrix}\text{ (option A)} \end{gathered}[/tex]
