Using cramer's Rule;
( 50 108 127 0 0.1 5.5 22 25.5 18 ) ( x y z ) = ( 393 5.7 91 ) \begin{pmatrix}
50&108& 127\\
0&0.1 & 5.5\\
22&25.5&18
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=\begin{pmatrix}
393 \\
5.7\\
91
\end{pmatrix} ⎝ ⎛ 50 0 22 108 0.1 25.5 127 5.5 18 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 393 5.7 91 ⎠ ⎞
Let D = D= D = ( 50 108 127 0 0.1 5.5 22 25.5 18 ) \begin{pmatrix}
50&108 & 127 \\
0 & 0.1&5.5\\
22&25.5&18
\end{pmatrix} ⎝ ⎛ 50 0 22 108 0.1 25.5 127 5.5 18 ⎠ ⎞
D x = ( 393 108 127 5.7 0.1 5.5 91 25.5 18 ) D_x=\begin{pmatrix}
393& 108&127\\
5.7&0.1 & 5.5\\
91&25.5&18
\end{pmatrix} D x = ⎝ ⎛ 393 5.7 91 108 0.1 25.5 127 5.5 18 ⎠ ⎞
D y = ( 50 393 127 0 5.7 5.5 22 91 18 ) D_y=\begin{pmatrix}
50&393 & 127 \\
0 & 5.7&5.5\\
22&91&18
\end{pmatrix} D y = ⎝ ⎛ 50 0 22 393 5.7 91 127 5.5 18 ⎠ ⎞
D 2 = ( 50 108 393 0 0.1 5.7 22 25.5 91 ) D_2=
\begin{pmatrix}
50&108 & 393 \\
0&0.1 & 5.7\\
22&25.5&91
\end{pmatrix} D 2 = ⎝ ⎛ 50 0 22 108 0.1 25.5 393 5.7 91 ⎠ ⎞
Det D = 50 ∣ 0.1 5.5 25.5 18 ∣ − 108 ∣ 0 5.5 22 18 ∣ + 127 ∣ 0 0.1 22 25.5 ∣ D=50\begin{vmatrix}
0.1 & 5.5 \\
25.5 & 18
\end{vmatrix}-108\begin{vmatrix}
0& 5.5 \\
22 & 18
\end{vmatrix}+127\begin{vmatrix}
0 & 0.1\\
22 & 25.5
\end{vmatrix} D = 50 ∣ ∣ 0.1 25.5 5.5 18 ∣ ∣ − 108 ∣ ∣ 0 22 5.5 18 ∣ ∣ + 127 ∣ ∣ 0 22 0.1 25.5 ∣ ∣
= 50 ( − 138.45 ) − 108 ( − 121 ) + 127 ( − 2.2 ) =50(-138.45)-108(-121)+127(-2.2) = 50 ( − 138.45 ) − 108 ( − 121 ) + 127 ( − 2.2 )
= 5866.1 =5866.1 = 5866.1
Det D x = 393 ∣ 0.1 5.5 25.5 18 ∣ − 108 ∣ 5.7 5.5 91 18 ∣ + 127 ∣ 5.7 0.1 91 25.5 ∣ D_x=393\begin{vmatrix}
0.1& 5.5 \\
25.5& 18
\end{vmatrix}-108\begin{vmatrix}
5.7& 5.5 \\
91& 18
\end{vmatrix}+127\begin{vmatrix}
5.7 & 0.1\\
91 & 25.5
\end{vmatrix} D x = 393 ∣ ∣ 0.1 25.5 5.5 18 ∣ ∣ − 108 ∣ ∣ 5.7 91 5.5 18 ∣ ∣ + 127 ∣ ∣ 5.7 91 0.1 25.5 ∣ ∣
= 393 ( − 138.45 ) − 108 ( − 397.9 ) + 127 ( 136.25 ) =393(-138.45)-108(-397.9)+127(136.25) = 393 ( − 138.45 ) − 108 ( − 397.9 ) + 127 ( 136.25 )
= 5866.1 =5866.1 = 5866.1
Det D y = 50 ∣ 5.7 5.5 91 18 ∣ − 393 ∣ 0 5.5 22 18 ∣ + 127 ∣ 0 5.7 22 91 ∣ D_y=50\begin{vmatrix}
5.7 & 5.5 \\
91& 18
\end{vmatrix}-393\begin{vmatrix}
0& 5.5 \\
22& 18
\end{vmatrix}+127\begin{vmatrix}
0& 5.7 \\
22 & 91
\end{vmatrix} D y = 50 ∣ ∣ 5.7 91 5.5 18 ∣ ∣ − 393 ∣ ∣ 0 22 5.5 18 ∣ ∣ + 127 ∣ ∣ 0 22 5.7 91 ∣ ∣
= 50 ( − 397.9 ) − 393 ( − 121 ) + 127 ( − 125.4 ) =50(-397.9)-393(-121)+127(-125.4) = 50 ( − 397.9 ) − 393 ( − 121 ) + 127 ( − 125.4 )
= 11732.2 =11732.2 = 11732.2
Det D 2 = 50 ∣ 0.1 5.7 25.5 91 ∣ − 108 ∣ 0 5.7 22 91 ∣ + 393 ∣ 0 0.1 22 25.5 ∣ D_2=50\begin{vmatrix}
0.1 & 5.7 \\
25.5 & 91
\end{vmatrix}-108\begin{vmatrix}
0& 5.7 \\
22 & 91
\end{vmatrix}+393\begin{vmatrix}
0& 0.1\\
22& 25.5
\end{vmatrix} D 2 = 50 ∣ ∣ 0.1 25.5 5.7 91 ∣ ∣ − 108 ∣ ∣ 0 22 5.7 91 ∣ ∣ + 393 ∣ ∣ 0 22 0.1 25.5 ∣ ∣
= 50 ( − 136.25 ) − 108 ( − 125.4 ) + 393 ( − 22 ) =50(-136.25)-108(-125.4)+393(-22) = 50 ( − 136.25 ) − 108 ( − 125.4 ) + 393 ( − 22 )
= 5866.1 =5866.1 = 5866.1
x = ∣ D x ∣ ∣ D ∣ = 5866.1 5866.1 = 1 , x=\frac{|D_x|}{|D|}=\frac{5866.1}{5866.1}=1, x = ∣ D ∣ ∣ D x ∣ = 5866.1 5866.1 = 1 ,
y = ∣ D y ∣ ∣ D ∣ = 11732.2 5866.1 = 2 y=\frac{|D_y|}{|D|}=\frac{11732.2}{5866.1}=2 y = ∣ D ∣ ∣ D y ∣ = 5866.1 11732.2 = 2
z = ∣ D z ∣ ∣ D ∣ = 5866.1 5866.1 = 1 z=\frac{|D_z|}{|D|}=\frac{5866.1}{5866.1}=1 z = ∣ D ∣ ∣ D z ∣ = 5866.1 5866.1 = 1
Amount of each brand of cereal that will give the level of nutrition required
Cereal Brand X = 1 X=1 X = 1 ounce
Cereal Brand Y = 2 Y=2 Y = 2 ounces
Cereal Brand Z = 1 Z=1 Z = 1 ounce
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