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Matrices
Cramers Rule

Probability Introduction :

1Q) Solve using Cramers Rule .

        x-2y+3z=7 ; 2x+y+z=4 ; -3x+2y-2z=-10 .

Solution :

      x-2y+3z=7 

     2x+y+z=4 

    -3x+2y-2z=-10

\(A= \begin{bmatrix}
1 & -2 &3 \\
 2& 1 &1 \\
 -3&2  &-2
\end{bmatrix}\)

\(X= \begin{bmatrix}
x\\
y\\

z\end{bmatrix} B=\begin{bmatrix}
7\\
4\\

-10\end{bmatrix}

\)

D=|A| =\(\begin{vmatrix}
1 &-2  & 3\\
 2& 1 &1 \\
-3 &2  & -2
\end{vmatrix}\)


D=\(1\begin{vmatrix}
1 & 1\\
 2&-2
\end{vmatrix}-(-2)\begin{vmatrix}
2 & 1\\
-3 &-2
\end{vmatrix}+3\begin{vmatrix}
2 & 1\\
-3 & 2
\end{vmatrix}\)

D=\(1(-2-2)+2(-4+3)+3(4+3)\)

D=-4-2+21=15

\(D_{x}=\begin{vmatrix}
7 &  -2&3 \\
 4& 1 &1 \\
 -10& 2 & -2
\end{vmatrix}
\)

\(D_{x}=30
\)
\(D_{y}= \begin{vmatrix}
1 &7  &3 \\
2 & 4 & 1\\
 -3& -10 & -2
\end{vmatrix}= -15\)

\(D_{z}=\begin{vmatrix}
1 & -2 &7 \\
 2& 1 &4 \\
 -3& 2 & -10
\end{vmatrix}=15\)

\(x= \frac{D_{x}}{D}=\frac{30}{15}=2\)

\(y= \frac{D_{y}}{D}=\frac{-15}{15}=-1\)

\(z= \frac{D_{z}}{D}=\frac{15}{15}=1\)

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