# 4 ?'s Help Me Please!!!

• Jul 24th 2006, 03:36 PM
Lane
questions on attachment!!
• Jul 24th 2006, 07:21 PM
CaptainBlack
Quote:

Originally Posted by Lane
questions on attachment!!

1. Find the 3 cofactors of the second row of:

$
A=\left [ \begin{array}{ccc}
{-7}& {-5}&{6}\\
{9}& {4}&{-3}\\
{8}&{-1}&{2}
\end{array} \right]
$

The cofactor $C_{i,j}$ is $(-1)^{i+j}$ times the determinant of the matrix obtained from $A$ by deleting the $i$th row
and the $j$th column of $A$.

The question is asking for $C_{2,1},\ C_{2,2}$ and $C_{2,3}$.

$
C_{2,1}=(-1)^{2+1}\
\left | \begin{array}{cc}
{-5}& {6}\\
{-1}& {2}
\end{array} \right|=-1[(-5)\times 2-(-1)\times 6]=4
$

$
C_{2,2}=(-1)^{2+2}\
\left | \begin{array}{cc}
{-7}& {6}\\
{8}& {2}
\end{array} \right|=\ ?$

$
C_{2,3}=\ ?$
,

where $?$ denotes something left to the reader to complete.

RonL
• Jul 24th 2006, 07:56 PM
Soroban
Hello, Lane1

You could help us by indicating where your difficulty lies.
Otherwise, we're not helping you ... we're just doing your homework.

Quote:

1. Find the three cofactors of the second row of the matrix.

. . . . $\begin{pmatrix}\text{-}7 & \text{-}5 & 6 \\ 9 & 4 & \text{-}3 \\ 8 & \text{-}1 & 2\end{pmatrix}$

Cofactor of $9:\;\;\begin{vmatrix}\text{-}5 & 6\\\text{-}1&2\end{vmatrix}$

Cofactor of $4:\;\;\begin{vmatrix}\text{-}7&6\\8&2\end{vmatrix}$

Cofactor of $\text{-}3:\;\;\begin{vmatrix}\text{-}7&\text{-}5\\8&\text{-}1\end{vmatrix}$

Quote:

2. Find the determinant of: $\begin{vmatrix}4&3&2\\3&2&4\\1&3&1\end{vmatrix}$
. . . $4\begin{vmatrix}2&4\\3&1\end{vmatrix} - 3\begin{vmatrix}5&4\\1&1\end{vmatrix} + 2\begin{vmatrix}5&2\\1&3\end{vmatrix}$ $\;= \;4(2-12) - 3(5-4) + 2(15-2)$

. . . $= \;4(-10) - 3(1) + 2(13) \;= \;-40 - 3 + 26 \;= \; \boxed{-17}$

Quote:

3. Use Cramer's Rule to solve (if possible) the system of equations.

. . . . $\begin{array}{cc}4x + 4y \:= \:1\\x + 5y \:=\:\text{-}4\end{array}$

$D\:=\:\begin{vmatrix}4&4\\1&5\end{vmatrix} \:= \:20 - 4 \:= \:16$

$D_x\:=\:\begin{vmatrix}1&4\\-4&5\end{vmatrix}\:=\:5+16\:=$ $\:21\quad\Rightarrow\quad \boxed{x \,= \,\frac{21}{16}}$

$D_y\:=\:\begin{vmatrix}4&1\\1&\text{-}4\end{vmatrix}\:=\:\text{-}16 - 1\:=\:-17\quad\Rightarrow\quad \boxed{y \,= \,-\frac{17}{16}}$

Quote:

the vertices of a triangle are given below. .Use a determinant
and the vertices of the triangle to find the area of the triangle.
. . . . . $(-2.5,\;3)\;\;(3,\;7.5)\;\;(5,\;5)$

The area of a triangle with vertices $(x_1,y_1),\;(x_2,y_2),\;(x_3,y_3)$ is given by:

. . . $A \;= \;\frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}$

So we have: . $A \;= \;\frac{1}{2}\begin{vmatrix}\text{-}2.5 & 3 & 1 \\ 3 & 7.5 & 1 \\ 5 & 5 & 1\end{vmatrix}$

I'll let you crank it out . . .

• Jul 24th 2006, 08:02 PM
CaptainBlack
2. Find the determinant of:

$
A=\left [ \begin{array}{ccc}
{4}& {3}&{2}\\
{5}& {2}&{4}\\
{1}&{3}&{1}
\end{array} \right]
$

Let's do this by expanding about the first row:

$
|A|=A_{1,1}\ C_{1,1}+A_{1,2}\ C_{1,2}+A_{1,3}\ C_{1,3}=$
$
4\ \left | \begin{array}{cc}
{2}& {4}\\
{3}& {1}
\end{array} \right|-$
$
3\ \left | \begin{array}{cc}
{5}& {4}\\
{1}& {1}
\end{array} \right|+
2\ \left | \begin{array}{cc}
{5}& {2}\\
{1}& {3}
\end{array} \right|
$
,

where $C_{i,j}$ is the $i,j$th cofactor od $A$

RonL