Hi guys

I came across this Inverse of matrices by reduction question and need you help(Surprised).

Find the inverse of the matrix A= [row1(1 2) row2(5 3)] and use it to solve { x+2y=7, 5x+3y=28 ??????

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- Oct 13th 2012, 08:55 PMprosri29Matrices anyone

Hi guys

I came across this Inverse of matrices by reduction question and need you help(Surprised).

Find the inverse of the matrix A= [row1(1 2) row2(5 3)] and use it to solve { x+2y=7, 5x+3y=28 ??????

- Oct 13th 2012, 09:02 PMAssassin0071Re: Matrices anyone
What you have to do is augment A to the identity matrix ( [A | I ] ) and then use elementary row operations until you turn A in to I. By doing these row operation on I in the same order you preformed them on A you will end up with the inverse of A (that is [A | I] turns in to [I | A^-1] after using elementary row operations.) Then you can use the inverse to solve the system (turn the system in to its matrix form then multiply both sides by the inverse). I hope that made sense (and helps).

- Oct 13th 2012, 09:20 PMprosri29Re: Matrices anyone
But I'm struggling to apply the proper elementary row operations Assassim0071 do you know the sequence of operations i could possibly use??

- Oct 13th 2012, 09:37 PMAssassin0071Re: Matrices anyone
Okay so you have a 2X2 matrix. to reduce that in to the identity you want to get 1's on the diagonals and zero's everywhere else. We can use row operation III (add a multiple of one row to another) to get turn the first entry of the second row to zero by adding -5 times the first row to the second. This elminates the 5 from the second row and turns the 3 in to a -7. Then you can use row operation II (multiply a row by a scalar) to turn the -7 in to a 1. Now we have 1's on the diagonal and only one other non zero entry, all we need to do is get rid of the 2 in the first row. We do this by using row operation III again and add -2 times the second row to the first. This should give you the identiy. So by preforming those exact operations on the identity you should obtain A inverse.