Math help solve using matrix equation

solve using matrix equation please
4x-2y=-12
3x-y= -3

4)

[ 4. . .-2 . . . -12 ------> 3R1 - 4R2 ----> R2
3. . .-1 . . . -3 ]

[ 4. . .-2 . . . -12 ------> R1 - R2 ----> R1
0. . .-2 . . . -24 ]

[ 4. . ..0 . . . .12 ------> R1/4 ----> R1
0. . .-2 . . . -24 ] R2/-2 ----> R2

[ 1. . ..0 . . . .3
0. . .1 . . . 12 ]

x = 3 & y = 12

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We can re-write this system in matrix form as follows:
[ 4 -2 ] [ x ] = [ -12 ]
[ 3 -1 ] [ y ]. . .[ -3 ].

Notice that the 2x2 matrix on the left contains the coefficients of x and y in the two equations, and the 2x1 matrix on the right contains the constants on the right side in the two equations.
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Now, call the 2x2 matrix on the left A, the 2x1 matrix on the LEFT X, and the 2x1 matrix on the right B. Then, we can re-write the equation as:
AX = B.

Multiplying both sides of this equation by A's inverse, A^-1, gives:
A^-1 * AX = A^-1 * B ==> X = A^-1 * B.
(note: A^-1 * AX = (A^-1 * A) * X = I * X = X, where I is the 2x2 identity matrix.)

By the formula for the inverse of a 2x2 matrix, we see that A has inverse:
A^-1 = [ -1/2 1 ]
. . . . . . [ -3/2 2 ].

Therefore, since A^-1 * B = [ 3 ]
. . . . . . . . . . . . . . . . . . . . . . [ 12 ]:
[ x ] = [ 3 ]
[ y ]. . [ 12 ] ==> x = 3 and y = 12, by comparing matrix elements.