Finding the inverse and finding a matrix * A = 0 matrix

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The discussion focuses on finding the inverse of a given matrix A and determining a matrix C that, when multiplied by another matrix B, results in a zero matrix. The user initially attempts to calculate the inverse of matrix A but makes errors in determining the determinant and applying the inverse formula. Clarifications are provided on the correct method to find the inverse, emphasizing the need to switch specific elements and divide by the determinant. Additionally, the concept of a null space is introduced while solving for matrix C, highlighting the importance of setting up the correct matrix product to achieve the desired zero matrix outcome. The conversation underscores the complexities involved in matrix operations and the necessity for accurate calculations.
mr_coffee
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Hello everyone! Matrices fun! Anyways,
I'm trying to find a matrix that is equal to the idenity matrix if u multiply 2 matrices together. Well that matrix is the inverse So I'm trying to find the inverse but the answer is wrong:
A =
-5 3
2 -9

So i found the determinant and i switched the a and d, and negated the c and b . isn't htat the inverse?

1/35-6 = 1/29


-9 -3
-2 -5

-9/29 -3/29
-2/29 -5/29

isn't thtat the inverse of A?

I also have B =
-1 6
5 -30

I need to multiply that by some matrix C so the resultant matrix is
0 0
0 0
but i can't just say C is equal to
0 0
0 0
any ideas? thanjks!

I did notice, Row 1 is just 1/5 row 2
 
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mr_coffee said:
So i found the determinant and i switched the a and d, and negated the c and b . isn't htat the inverse?
You have to switch a and d, also switch b and c and then change the signs of these last 2. After that, you have to divide by the determinant. So:

A = \left( {\begin{array}{*{20}c}<br /> a &amp; b \\<br /> c &amp; d \\<br /> <br /> \end{array} } \right) \Rightarrow A^{ - 1} = \frac{1}<br /> {{\det A}}\operatorname{adj} A = \frac{1}<br /> {{ad - bc}}\left( {\begin{array}{*{20}c}<br /> d &amp; { - c} \\<br /> { - b} &amp; a \\<br /> <br /> \end{array} } \right)

mr_coffee said:
I need to multiply that by some matrix C so the resultant matrix is
0 0
0 0
but i can't just say C is equal to
0 0
0 0
any ideas? thanjks!
If you don't "see" it right away, work out the following matrix product to get a system of lineair equations.

\left( {\begin{array}{*{20}c}<br /> { - 1} &amp; 6 \\<br /> 5 &amp; { - 30} \\<br /> <br /> \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br /> a &amp; b \\<br /> c &amp; d \\<br /> <br /> \end{array} } \right) = \left( {\begin{array}{*{20}c}<br /> 0 &amp; 0 \\<br /> 0 &amp; 0 \\<br /> <br /> \end{array} } \right)
 
Multiplying a matrix by something to get zero strongly reminds me of the notion of a null space...
 
Thanks TD, but for some reason its still wrong, i got:
-9/29 -2/29
-3/29 -5/29
 
Check your determinant again, that is ad-bc :smile:
 
lol what the f, (-5)(-9) - (3)(2)
1/45-6
1/39 right?
 
Well, the determinant is 39 (so not 29). Then, you have to divide by it indeed.
So 1/39 * adj(A)
 

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