Matrix Inverse Problem: Troubleshooting Proof and Multiplication Error

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The discussion revolves around a matrix inverse problem where the user is confused about the multiplication of a matrix and its inverse yielding the identity matrix. They initially believe that the multiplication does not equal 1, specifically questioning the calculation of -10/2 + 6. However, others clarify that the multiplication of the matrices indeed results in the identity matrix, confirming the user's calculations were incorrect. The correct computation shows that the product does equal 1 for the corresponding elements, validating the proof of the inverse. The conversation emphasizes the importance of double-checking matrix multiplication calculations.
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im doing a matrix inverse problem and have the inverse but when i do the proof the book says it should look like this:

2 -3 -5/2 -3/2 equals 1 0
-4 5 times -2 -1 0 1

but if you muliply those shouldn't you get -10/2 + 6 which doesn't equal 1

help please
 
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What do you mean by those numbers? Do you mean this:

\left(\begin{array}{cc}2&-3\\-4&5\end{array}\right)\times\left(\begin{array}{cc}-5/2&-3/2\\-2&-1\end{array}\right)

If so, that does indeed give you the identity matrix. Check your calculation again.
 
DethRose said:
im doing a matrix inverse problem and have the inverse but when i do the proof the book says it should look like this:

2 -3 -5/2 -3/2 equals 1 0
-4 5 times -2 -1 0 1

but if you muliply those shouldn't you get -10/2 + 6 which doesn't equal 1

help please
As a matter of fact, it does:
\left( \begin{array}{cc} 2 & -3 \\ -4 & 5 \end{array} \right) \times \left( \begin{array}{cc} -\frac{5}{2} & -\frac{3}{2} \\ -2 & -1 \end{array} \right) = \left( \begin{array}{cc} 2 \times \left( -\frac{5}{2} \right) + (-3) \times (-2) & 2 \times \left( -\frac{3}{2} \right) + (-3) \times (-1) \\ -4 \times \left( -\frac{5}{2} \right) + 5 \times (-2) & -4 \times \left( -\frac{3}{2} \right) + 5 \times (-1) \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), the desired result. :)
 
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DethRose said:
but if you muliply those shouldn't you get -10/2 + 6 which doesn't equal 1
-10/2 + 6 = -5 + 6 = 6 - 5 = ?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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