Finding the Determinant of a Transformed Matrix

[nicknaq]

Homework Statement

A=(r1, r2, r3)transpose and detA=2. Let B=(2r1+3r2, r3+2r1, r2-r1-3r3). Find detB.

Homework Equations

- for row switch the det is multiplied by -1
- if you add a row to another it has no effect on det
- if you multiply a row by a constant k the det will be k*det

The Attempt at a Solution

I broke matrix B up as follows:
(2r1, r3, r2) + (3r2, 2r1, -3r3-r1)

Then the det will be (2)(-1)(2)+(-1)(3)(2)(-3)= 32.
Can someone verify my work? I think it's wrong.

 

[vela]

It's wrong. It's not true that det(M+N)=det(M)+det(N).

 

[nicknaq]

It's wrong. It's not true that det(M+N)=det(M)+det(N).

Thanks. I thought it was.
Can you help me solve it?

 

[vela]

First, how are det(A) and det(A^T) related? Once you know det(A^T), you can apply row operations to A^T to transform it into B. By keeping track of those operations, you can deduce how det(B) is related to det(A).

 

[nicknaq]

First, how are det(A) and det(A^T) related? Once you know det(A^T), you can apply row operations to A^T to transform it into B. By keeping track of those operations, you can deduce how det(B) is related to det(A).

Shoot. The question is written wrong. My apologies.
Both matrices are transpose. I.e. they both take this form

| r1 |
| r2 |
| r3 |
| r4 |

Again sorry. I still very much need help though.

 

[vela]

That's even simpler then. Just see what row operations you have to apply to A to turn it into B.

 

[nicknaq]

going from:
| r1 |
| r2 |
| r3 |
to
|2r1+3r2|
|r3+2r1|
|r2-r1-3r3|

I would need to do the following
r1*2 --> changes det by *2
+3r2 to r1 --> no change on det
switch r3 and r2 --> det gets a negative sign
+2r1 to r2 --> no change on det?
-r1-3r3 to r3--> no change on det?

 

[vela]

Sorry, I just realized that's not going to work easily because the row operations change the rows. So, for example, the elementary operations of multiplying row 1 by 2 and followed by adding row 1 to row 3 results in the sequence

<br /> \begin{vmatrix}r1\\r2\\r3\end{vmatrix} \rightarrow \begin{vmatrix}2r1\\r2\\r3\end{vmatrix} \rightarrow<br /> \begin{vmatrix}r1\\r2\\r3+2r1<br /> \end{vmatrix}<br />

You don't end up at

<br /> \begin{vmatrix}2r1\\r2\\r3+r1\end{vmatrix}<br />

as you might expect. If you keep track of what's in the new rows and how to combine them in just the right way to get the final matrix you want, this method will work, but there's a simpler way. The most straightforward way to solve this problem is to use the fact that the determinant is a linear function of its rows. What this means, for example, is that

\begin{vmatrix}2r1+3r2\\r3+2r1\\r2-r1-3r3\end{vmatrix} = 2\begin{vmatrix}r1\\r3+2r1\\r2-r1-3r3\end{vmatrix} + 3\begin{vmatrix}r2\\r3+2r1\\r2-r1-3r3\end{vmatrix}

Now if you work on the second row of the first matrix on the RHS, you have

\begin{vmatrix}r1\\r3+2r1\\r2-r1-3r3\end{vmatrix} = 2\begin{vmatrix}r1\\r1\\r2-r1-3r3\end{vmatrix} + \begin{vmatrix}r1\\r3\\r2-r1-3r3\end{vmatrix} = \begin{vmatrix}r1\\r3\\r2-r1-3r3\end{vmatrix}

The matrix with r1 as both its first and second row has a determinant of 0, so you can get rid of it. Eventually you'll get to matrices with determinant 0 or matrices consisting of some permutation of r1, r2, and r3, whose determinant will be either +det(A) or -det(A). When you simplify at the end, you should find det(B)=13 det(A).