MHB 311.3.1.1 - Determinants And Cofactor Expansion

[karush]

nmh{898}
311 Determinants And Cofactor Expansion (3.1.1)

a. Compare the determinants using a cofactor expansion across the first row.

b. compute the determinant by a cofactor expansion down the second column.

$$\left|
\begin{array}{rrr}
3&0& 4\\
2&3& 2\\
0&5&-1\\
\end{array}
\right|$$

ok I accually start this class tomorro but thot I would try some basic stuff
already stuck...

 

[skeeter]

Re: 311.3.1.1

a. Compare the determinants using a cofactor expansion across the first row.

b. compute the determinant by a cofactor expansion down the second column.

$$\left|
\begin{array}{rrr}
3&0& 4\\
2&3& 2\\
0&5&-1\\
\end{array}
\right|$$

ok I accually start this class tomorro but thot I would try some basic stuff
already stuck...

(a)

$3\begin{vmatrix}
3 & 2\\
5 & -1
\end{vmatrix} - 0\begin{vmatrix}
2 &2 \\
0 & -1
\end{vmatrix}+4\begin{vmatrix}
2 & 3\\
0& 5
\end{vmatrix} =3(-3-10)-0(-2-0)+4(10-0)=1$

(b)

note $\begin{vmatrix}
3 & 0 & 4\\
2 & 3 & 2\\
0 & 5& -1
\end{vmatrix}=\begin{vmatrix}
0 & 4 & 3\\
3 & 2 & 2\\
5 & -1 & 0
\end{vmatrix}$

the second determinant is formed by moving the 1st column in the original matrix to the 3rd column. Now use the new 1st column and its cofactors ...

$0\begin{vmatrix}
2 & 2\\
-1 & 0
\end{vmatrix}-3\begin{vmatrix}
4 & 3\\
-1 & 0
\end{vmatrix}+5\begin{vmatrix}
4 &3 \\
2 & 2
\end{vmatrix}=0[0-(-2)]-3[0-(-3)]+5(8-6)=1$

This is a method I learned in undergrad physics that "stuck". There are other methods to do these ... I'm sure someone else well-versed in linear algebra will contribute.

 

[MarkFL]

Please give threads titles that briefly describe the thread topic. A thread title that is simply a series of digits and periods is not what we consider a good thread title. :)

 

[karush]

https://dl.orangedox.com/GXEVNm73NxaGC9F7Cy