# Math Help - 4 ?'s Help Me Please!!!

1. ## 4 ?'s Help Me Please!!!

questions on attachment!!

2. 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

3. Hello, Lane1

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

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

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

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

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

. . . . $\begin{array}{cc}4x + 4y \:= \:1\\x + 5y \:=\:\text{-}4\end{array}$
$D\:=\:\begin{vmatrix}4&4\\1&5\end{vmatrix} \:= \:20 - 4 \:= \:16$

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

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

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 . . .

4. 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