My Math Forum How to find inverse of 3 by 2 matrix dont use echelon form

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 December 13th, 2014, 09:27 PM #1 Newbie   Joined: Dec 2014 From: pakistan Posts: 15 Thanks: 0 How to find inverse of 3 by 2 matrix dont use echelon form hi, how to find inverse of 3 by 2 matrix.dont use echelon form,reduced echelon form.use cofactors method.
 December 14th, 2014, 08:51 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 You don't, as a 3 by 2 matrix isn't a square matrix.
 December 15th, 2014, 06:00 AM #3 Newbie   Joined: Dec 2014 From: pakistan Posts: 15 Thanks: 0 How to find inverse of 3 by 2 matrix dont use echelon form to see the real question plz click link below https://www.dropbox.com/s/ug9e2n702j...atrix.bmp?dl=0 in this question find value of matrix A. Last edited by ormara; December 15th, 2014 at 06:03 AM. Reason: revised post
 December 15th, 2014, 06:09 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 A is a 2*2 matrix. Assign a letter to each of the four elements, multiply and see whether it has a solution.
 December 16th, 2014, 12:31 AM #5 Newbie   Joined: Dec 2014 From: pakistan Posts: 15 Thanks: 0 How to find inverse of 3 by 2 matrix dont use echelon form did exactly the same but the book's answer was different than mine.how can you say A is a 2 by 2 matrix.it can be 3 by 3,1 by 3,3 by 4,3 by 2 etc. (off topic)why my thread was moved. Last edited by ormara; December 16th, 2014 at 12:39 AM. Reason: adding more comments
 December 16th, 2014, 04:50 AM #7 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions $\displaystyle A$ must be a 2 x 2 matrix because if it didn't the matrix on the right-hand side would have different dimensions. $\displaystyle \left[\begin{array}[cc] \\ 5 & -1 \\ 0 & 0 \\ 3 & 1 \\ \end{array}\right] \left[\begin{array}[cc] \\ A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{array}\right] = \left[\begin{array}[cc] \\ 3 & -7 \\ 0 & 0 \\ 7 & 2 \\ \end{array}\right]$ As Hoempa suggested, we multiply out the matrices to get the following simultaneous equations: $\displaystyle 5A_{11} - A_{21} = 3$ $\displaystyle 5A_{12} - A_{22} = -7$ $\displaystyle 3A_{11} + A_{21} = 7$ $\displaystyle 3A_{12} + A_{22} = 2$ The solutions to these are $\displaystyle A_{11} = \frac{5}{4}$ $\displaystyle A_{12} = \frac{-5}{8}$ $\displaystyle A_{21} = \frac{13}{4}$ $\displaystyle A_{22} = \frac{31}{8}$ So $\displaystyle A = \left[\begin{array}[cc] \\ A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{array}\right] = \frac{1}{8} \left[\begin{array}[cc] \\ 10 & -5 \\ 28 & 31 \\ \end{array}\right]$ Last edited by Benit13; December 16th, 2014 at 04:53 AM.
 December 16th, 2014, 05:07 AM #8 Newbie   Joined: Dec 2014 From: pakistan Posts: 15 Thanks: 0 @Benit13 order formula is if we have a matrix equation A x B = C A has order m by n, B has order n by p, then its product A x B has order m by p.This formula is in textbooks.so by this formula answer should have order 3 by 2 but answer has order 2 by 2.

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