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Use the inverse A^(-1) in (a) to solve Ax.
Ax=
[2]
[1]
[3]
the inverse for (a) is; (**dots just for spacing)
[1......-3......2]
[1/2....1...-1/2]
[-1.....-1......1]
answer should be
[1]
[1/2]
[0]
can someone explain how this answer is obtained. 10pts for explanation
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Answers (3)
The top row of your matrix is incorrect. If Ax = b, and if A has an inverse, then the solution vector x is given by A^(-1)b. So you should only have to multiply your inverse matrix by the column (2,1,3).
This is done by multiplying the elements of the row of the matrix with the entries in the column and adding. So for the matrix you are listing as A^(-1), the product gives
[1(2) + (-3)(1) + 2(3)]
[½(2) + 1(1) + (-½)(3)]
[-1(2)+(-1)(1) + 1(3)]
But if you work that out, you get
[5]
[½]
[0]
The 1 in the first entry of your solution should be a 5. Are you sure there isn't a typo in your post?
The numbers you have given do not agree with the answer you gave.
This method of solution is based on the fact that if
AX = B where A is the matrix of coefficients and B is the matrix of constants, then
A^-1AX = A^-1B and A^-1A = I where I is the identity matrix so
X = A^-1B
The numbers you have given for A^-1 and B yield x = 5, y = 1/2, and z = 0.
Just multiply Ax by A^-1 on the left.
Reference: Verified on TI-84 calculator.