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

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    Inverse Matrix
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Homework Help Overview

The discussion revolves around finding the inverse of a given matrix and determining a matrix that, when multiplied by another, results in a zero matrix. The subject area includes linear algebra, specifically matrix operations and properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the inverse of a matrix by calculating its determinant and applying the formula for the inverse. They express uncertainty about their results and seek confirmation. Another participant provides clarification on the correct method for finding the inverse, including the need to divide by the determinant. Additionally, the original poster seeks guidance on finding a matrix that results in a zero matrix when multiplied by another matrix, noting a relationship between the rows of the second matrix.

Discussion Status

Participants are actively engaging with the problem, providing insights and corrections regarding the calculation of the determinant and the process of finding the inverse. There is a recognition of the need to clarify the steps involved in these calculations, and some participants are exploring the implications of matrix multiplication leading to a zero matrix.

Contextual Notes

There is a mention of a potential misunderstanding regarding the determinant calculation, with participants questioning the values derived from their computations. The original poster also notes a specific relationship between the rows of one of the matrices, which may influence their approach.

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:

[tex]A = \left( {\begin{array}{*{20}c}<br /> a & b \\<br /> c & 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 & { - c} \\<br /> { - b} & a \\<br /> <br /> \end{array} } \right)[/tex]

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.

[tex]\left( {\begin{array}{*{20}c}<br /> { - 1} & 6 \\<br /> 5 & { - 30} \\<br /> <br /> \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br /> a & b \\<br /> c & d \\<br /> <br /> \end{array} } \right) = \left( {\begin{array}{*{20}c}<br /> 0 & 0 \\<br /> 0 & 0 \\<br /> <br /> \end{array} } \right)[/tex]
 
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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