Answer: Given: \textsf{Equations are}Equations are \mathsf{x+y+z=6,\;x-y=z=2,\;x+2y-z=2}x+y+z=6,x−y=z=2,x+2y−z=2 \underline{\textsf{To find:}} To find: \textsf{Solution by Cramer’s rule}Solution by Cramer’s rule \underline{\textsf{Solution:}} Solution: \begin{gathered}\mathsf{\triangle=\left|\begin{array}{ccc}1&1&1\\1&-1&1\\1&2&-1\end{array}\right|}\end{gathered} △= ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 1 1 −1 2 1 1 −1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ \mathsf{\triangle=1(1-2)-1(-1-1)+1(2+1)}△=1(1−2)−1(−1−1)+1(2+1) \mathsf{\triangle=-1+2+3=4}△=−1+2+3=4 \begin{gathered}\mathsf{{\triangle}_x=\left|\begin{array}{ccc}6&1&1\\2&-1&1\\2&2&-1\end{array}\right|}\end{gathered} △ x = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 6 2 2 1 −1 2 1 1 −1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ \mathsf{{\triangle}_x=6(1-2)-1(-2-2)+1(4+2)}△ x =6(1−2)−1(−2−2)+1(4+2) \mathsf{{\triangle}_x=-6+4+6=4}△ x =−6+4+6=4 \begin{gathered}\mathsf{{\triangle}_y=\left|\begin{array}{ccc}1&6&1\\1&2&1\\1&2&-1\end{array}\right|}\end{gathered} △ y = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 1 6 2 2 1 1 −1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ \mathsf{{\triangle}_y=1(-2-2)-6(-1-1)+1(2-2)}△ y =1(−2−2)−6(−1−1)+1(2−2) \mathsf{{\triangle}_y=-4+12+0=8}△ y =−4+12+0=8 \begin{gathered}\mathsf{{\triangle}_z=\left|\begin{array}{ccc}1&1&6\\1&-1&2\\1&2&\end{array}\right|}\end{gathered} △ z = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 1 1 1 −1 2 6 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ \mathsf{{\triangle}_z=1(-2-4)-1(2-2)+6(2+1)}△ z =1(−2−4)−1(2−2)+6(2+1) \mathsf{{\triangle}=-6+18=12}△=−6+18=12 \textsf{By cramer’s rule}By cramer’s rule \mathsf{x=\dfrac{\triangle_x}{\triangle}=\dfrac{4}{4}=1}x= △ △ x = 4 4 =1 \mathsf{y=\dfrac{\triangle_y}{\triangle}=\dfrac{8}{4}=2}y= △ △ y = 4 8 =2 \mathsf{z=\dfrac{\triangle_z}{\triangle}=\dfrac{12}{4}=3}z= △ △ z = 4 12 =3 \therefore\textsf{The solution is (x,y,z)=(1,2,3)}∴The solution is (x,y,z)=(1,2,3) Reply
Answer:
Given:
\textsf{Equations are}Equations are
\mathsf{x+y+z=6,\;x-y=z=2,\;x+2y-z=2}x+y+z=6,x−y=z=2,x+2y−z=2
\underline{\textsf{To find:}}
To find:
\textsf{Solution by Cramer’s rule}Solution by Cramer’s rule
\underline{\textsf{Solution:}}
Solution:
\begin{gathered}\mathsf{\triangle=\left|\begin{array}{ccc}1&1&1\\1&-1&1\\1&2&-1\end{array}\right|}\end{gathered}
△=
∣
∣
∣
∣
∣
∣
∣
1
1
1
1
−1
2
1
1
−1
∣
∣
∣
∣
∣
∣
∣
\mathsf{\triangle=1(1-2)-1(-1-1)+1(2+1)}△=1(1−2)−1(−1−1)+1(2+1)
\mathsf{\triangle=-1+2+3=4}△=−1+2+3=4
\begin{gathered}\mathsf{{\triangle}_x=\left|\begin{array}{ccc}6&1&1\\2&-1&1\\2&2&-1\end{array}\right|}\end{gathered}
△
x
=
∣
∣
∣
∣
∣
∣
∣
6
2
2
1
−1
2
1
1
−1
∣
∣
∣
∣
∣
∣
∣
\mathsf{{\triangle}_x=6(1-2)-1(-2-2)+1(4+2)}△
x
=6(1−2)−1(−2−2)+1(4+2)
\mathsf{{\triangle}_x=-6+4+6=4}△
x
=−6+4+6=4
\begin{gathered}\mathsf{{\triangle}_y=\left|\begin{array}{ccc}1&6&1\\1&2&1\\1&2&-1\end{array}\right|}\end{gathered}
△
y
=
∣
∣
∣
∣
∣
∣
∣
1
1
1
6
2
2
1
1
−1
∣
∣
∣
∣
∣
∣
∣
\mathsf{{\triangle}_y=1(-2-2)-6(-1-1)+1(2-2)}△
y
=1(−2−2)−6(−1−1)+1(2−2)
\mathsf{{\triangle}_y=-4+12+0=8}△
y
=−4+12+0=8
\begin{gathered}\mathsf{{\triangle}_z=\left|\begin{array}{ccc}1&1&6\\1&-1&2\\1&2&\end{array}\right|}\end{gathered}
△
z
=
∣
∣
∣
∣
∣
∣
∣
1
1
1
1
−1
2
6
2
∣
∣
∣
∣
∣
∣
∣
\mathsf{{\triangle}_z=1(-2-4)-1(2-2)+6(2+1)}△
z
=1(−2−4)−1(2−2)+6(2+1)
\mathsf{{\triangle}=-6+18=12}△=−6+18=12
\textsf{By cramer’s rule}By cramer’s rule
\mathsf{x=\dfrac{\triangle_x}{\triangle}=\dfrac{4}{4}=1}x=
△
△
x
=
4
4
=1
\mathsf{y=\dfrac{\triangle_y}{\triangle}=\dfrac{8}{4}=2}y=
△
△
y
=
4
8
=2
\mathsf{z=\dfrac{\triangle_z}{\triangle}=\dfrac{12}{4}=3}z=
△
△
z
=
4
12
=3
\therefore\textsf{The solution is (x,y,z)=(1,2,3)}∴The solution is (x,y,z)=(1,2,3)