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Question

Using matrices, solve the following system of equation.
2x3y+5z=11
3x+2y4z=5
x+y2z=3

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Solution

Let solution is X=xyz
Then, AX=B, where A=235324112 and B=1153
X=A1B
We first find A1,
Step 1: Calculate the Matrix of Minors, let it be M
M=02119522313
Step 2:then turn it into matrix of co-factors,
This is easy! Just apply a "checkerboard" of minuses to the "Matrix of Minors". In other words, we need to change the sign of alternate cells, like this:
M=02119522313 with +++++M=02119522313
Step 3:then the adjugate:
Now "Transpose" all elements of the previous matrix... in other words swap their positions over the diagonal (the diagonal stays the same):
M=01229231513
Step 4:Multiply by 1Determinant
Det(A)=1
M=1Determinant01229231513=01229231513
Hence A1=M
So X=MB=012292315131153
=0×11+1×5+(2)×(3)(2)×11+9×(5)+(23)×(3)(1)×11+5×(5)+(13)×(3)=123
x=1,y=2,z=3

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