g i v e n y − z = 2 3 x + 2 y + z = 4 5 x + 4 y = 1 w e c a n w r i t e t h i s a s 0 x + y − z = 2 3 x + 2 y + z = 4 5 x + 4 y + 0 z = 1 S t e p 1 : F i n d t h e d e t e r m i n a n t o f t h e c o e f f i c i e n t m a t r i x . ∣ 0 1 − 1 3 2 1 5 4 0 ∣ N o w f i n d d e t e r m i n a n t a s D = 0 ( 2 × 0 − 1 × 4 ) + 1 ( 1 × 5 − 3 × 0 ) + ( − 1 ) ( 3 × 4 − 2 × 5 ) = 3 S t e p 2 : F i n d t h e d e t e r m i n a n t o f t h e x − m a t r i x ( D x ) . X − m a t r i x i s f o r m e d b y r e p l a c i n g t h e x − c o l u m n v a l u e s w i t h t h e a n s w e r − c o l u m n v a l u e s ∣ 2 1 − 1 4 2 1 1 4 0 ∣ N o w f i n d d e t e r m i n a n t = D x = − 21 S t e p 3 : F i n d t h e d e t e r m i n a n t o f t h e y − m a t r i x ( D y ) . Y − m a t r i x i s f o r m e d b y r e p l a c i n g t h e y − c o l u m n v a l u e s w i t h t h e a n s w e r − c o l u m n v a l u e s ∣ 0 2 − 1 3 4 1 5 1 0 ∣ N o w f i n d d e t e r m i n a n t = D y = 27 S t e p 4 : F i n d t h e d e t e r m i n a n t o f t h e z − m a t r i x ( D z ) . Z − m a t r i x i s f o r m e d b y r e p l a c i n g t h e z − c o l u m n v a l u e s w i t h t h e a n s w e r − c o l u m n v a l u e s ∣ 0 1 2 3 2 4 5 4 1 ∣ N o w f i n d d e t e r m i n a n t = D z = 21 C r a m e r s R u l e s a y s t h a t t h e s o l u t i o n s a r e x = D x D = − 21 3 = − 7 y = D y D = 27 3 = 9 z = D z D = 21 3 = 7 given\\
y − z = 2\\
3x + 2y + z = 4\\
5x + 4y =1 \\
we\space can\space write\space this\space as\\
0x+y − z = 2\\
3x + 2y + z = 4\\
5x + 4y +0z=1 \\
Step \space 1: \space Find\space the \space determinant\space of \space the\space coefficient\space matrix.\\
\begin{vmatrix}
0 & 1 & -1\\
3 & 2 & 1\\
5 & 4 &0
\end{vmatrix} \\
Now\space find\space determinant \space as\space \\
D=0(2×0-1×4)+1(1×5-3×0)+(-1)(3×4-2×5)=3\\
Step2:\space Find \space the \space determinant\space of\space the\space x -\space matrix\space ( D_x ).\space X \space - \space matrix\space is \space formed\space by \space replacing\space the \space x-column\space values \space with \space the \space answer-column\space values\\
\begin{vmatrix}
2 & 1 & -1\\
4& 2 & 1\\
1 & 4 &0
\end{vmatrix} \\
Now \space find \space determinant \space = D_x =-21\\
Step3:\space Find\space the \space determinant\space of \space the\space y - matrix\space \space ( D_y ). \space Y - matrix\space is \space formed\space by \space replacing \space the\space y-column \space values\space with \space the \space answer-column \space values\\
\begin{vmatrix}
0 & 2& -1\\
3 & 4 & 1\\
5 & 1&0
\end{vmatrix} \\
Now\space find \space determinant \space = D_y= 27\\
Step4:\space Find\space the \space determinant\space of \space the\space z - matrix\space \space ( D_z ). \space Z - matrix\space is \space formed\space by \space replacing \space the\space z-column \space values\space with \space the \space answer-column \space values\\
\begin{vmatrix}
0 & 1 & 2\\
3 & 2& 4\\
5 & 4 &1
\end{vmatrix} \\
Now\space find \space determinant \space = D_z= 21\\
Cramers \space Rule\space says\space that\space the\space solutions \space are\\
x=\frac{D_x }{D}=\frac{-21}{3}=-7\\
y=\frac{D_y }{D}=\frac{27}{3}=9\\
z=\frac{D_z }{D}=\frac{21}{3}=7\\ g i v e n y − z = 2 3 x + 2 y + z = 4 5 x + 4 y = 1 w e c an w r i t e t hi s a s 0 x + y − z = 2 3 x + 2 y + z = 4 5 x + 4 y + 0 z = 1 St e p 1 : F in d t h e d e t er minan t o f t h e coe ff i c i e n t ma t r i x . ∣ ∣ 0 3 5 1 2 4 − 1 1 0 ∣ ∣ N o w f in d d e t er minan t a s D = 0 ( 2 × 0 − 1 × 4 ) + 1 ( 1 × 5 − 3 × 0 ) + ( − 1 ) ( 3 × 4 − 2 × 5 ) = 3 St e p 2 : F in d t h e d e t er minan t o f t h e x − ma t r i x ( D x ) . X − ma t r i x i s f or m e d b y re pl a c in g t h e x − co l u mn v a l u es w i t h t h e an s w er − co l u mn v a l u es ∣ ∣ 2 4 1 1 2 4 − 1 1 0 ∣ ∣ N o w f in d d e t er minan t = D x = − 21 St e p 3 : F in d t h e d e t er minan t o f t h e y − ma t r i x ( D y ) . Y − ma t r i x i s f or m e d b y re pl a c in g t h e y − co l u mn v a l u es w i t h t h e an s w er − co l u mn v a l u es ∣ ∣ 0 3 5 2 4 1 − 1 1 0 ∣ ∣ N o w f in d d e t er minan t = D y = 27 St e p 4 : F in d t h e d e t er minan t o f t h e z − ma t r i x ( D z ) . Z − ma t r i x i s f or m e d b y re pl a c in g t h e z − co l u mn v a l u es w i t h t h e an s w er − co l u mn v a l u es ∣ ∣ 0 3 5 1 2 4 2 4 1 ∣ ∣ N o w f in d d e t er minan t = D z = 21 C r am ers R u l e s a ys t ha t t h e so l u t i o n s a re x = D D x = 3 − 21 = − 7 y = D D y = 3 27 = 9 z = D D z = 3 21 = 7
#332078 on Oct 2022
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