Determinants of A and B

**al1850** · April 5th 2008, 01:07 PM

Find the Determinants of A and B.
Please show me how to do it,especially for B.
Appreciate your help.

**Mathstud28** · April 5th 2008, 01:21 PM

using latex would take forever...so I suggest you look up either the "alternate method" or "expansion by cofactors" method of finding determinants...I generally prefer the alternate method for 3X3 matricies for their simplicity...but to make sure you understand..the answer for B is 1

**galactus** · April 5th 2008, 01:22 PM

You can find the det, this way:

$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22} &a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$

$det=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_ {21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}$

I will use x instead of theta for less typing:

$(cosx)(cosx)(1)+sinx(0)(0)+(0)(-sinx)(0)-(0)(cosx)(0)$ $-(sinx)(-sinx)(1)-(cosx)(0)(0)=cos^{2}x+sin^{2}x=1$

**Moo** · April 5th 2008, 01:29 PM

Hello,

If you have a matrix :

$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$

Write :

$\begin{pmatrix} a & b & c & a & b \\ d & e & f & d & e \\ g & h & i & g & h \end{pmatrix}$

The determinant will be found by adding the product of terms in the diagonals from up-left to down-right and substracting the product of terms in the diagonals from up-right to down-left.

$\begin{pmatrix} \boxed{a} & \boxed{b} & \boxed{c} & a & b \\ d & \boxed{e} & \boxed{f} & \boxed{d} & e \\ g & h & \boxed{i} & \boxed{g} & \boxed{h} \end{pmatrix}$

aei+bfg+cdh

$\begin{pmatrix} a & b & \boxed{c} & \boxed{a} & \boxed{b} \\ d & \boxed{e} & \boxed{f} & \boxed{d} & e \\ \boxed{g} & \boxed{h} & \boxed{i} & g & h \end{pmatrix}$

-ceg-afh-bdi

The determinant is : aei+bfg+cdh-ceg-afh-bdi

**al1850** · April 5th 2008, 04:09 PM

thanks,you guys are always so helpful, no matter how small or simple my questions are ;-)

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