If Let "\\begin{pmatrix}\n 2 & 1 \\\\\n 5 & 3\n\\end{pmatrix}\\begin{pmatrix}\n x \\\\\ny\n\\end{pmatrix} =\n\\begin{pmatrix}\n 2 \\\\\n 4\n\\end{pmatrix}",
Solve for x and y
"\\begin{pmatrix} 2 & 1 \\\\ 5 & 3 \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix}"
Performing matrix multiplication
"\\begin{pmatrix}2 x+y \\\\5x+3 y \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix}"
Equating corresponding elements we get
2x + y = 2 ••••••••••(1)
5x + 3y = 4. ••••••••••(2)
Multiplying equation (1) by 3 we get
6x + 3y = 6 ••••••••••(3)
Then subtracting equation (2) from equation (3) we get
6x + 3y - 5x - 3y = 6 - 4
=> x = 2
Substituting x = 2 in equation (1) we get
4 + y = 2
=> y = 2 - 4 = - 2
So solution is x = 2 and y = -2
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