Linear Algebra: Help Solving a Matrix Equation?

Well I would take the inverse of the first matrix, the 2x2 one, and then multiply it by the matrix on the right hand side.

If AX=B

Then X = inv(A) times B

no longer between the above explanation: -x+y=-5 2x+y=-8 you will would prefer to subtract the two equations from one yet yet another to do away with a variable, subsequently, the y. you would be left with... -3x=3 Divide via using 3... -x=a million x=-a million then you replace the x interior the 1st equation with -a million... -(-a million)+y=-5 a million+y=-5 Transpose... y=-5-a million y=-6 So, interior the words of (x.y), the respond is (-a million,-6) or no longer between the above.

[ 2 5] [x] = [ 15]

[1 -1] [y] [4]

(coefficient m)*(variable m) =(constant m)

A * v = B

v = A^(-1) * B

then v = [5]

[1]

which means x = 5 and y = 1