solve 3x+3y-z=11,2x-y+2z=9,4x+3y+2z=25 by cramer's rule
Answers
Answer:
✵\textbf\red{Given:}Given: ✵
\mathsf{x+y+z=6}x+y+z=6
\mathsf{3x+3y+z=12}3x+3y+z=12
\mathsf{2x+3y+2z=14}2x+3y+2z=14
\textbf\red{To find:}To find:
\textsf{Solution\: of \:the\: given\: system \:of\: equations \:by \:Cramer's \:rule}Solutionof thegivensystem ofequations by Cramer’s rule
\textbf\red{Solution:}Solution:
\begin{gathered}\begin{gathered}\mathsf{\triangle=\left|\begin{array}{ccc}1&1&1\\3&3&1\\2&3&2\end{array}\right|}\end{gathered} \end{gathered}
△=
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1
3
2
1
3
3
1
1
2
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\mathsf{\triangle=1(6-3)-1(6-2)+1(9-6)}△=1(6−3)−1(6−2)+1(9−6)
\mathsf{\triangle=3-4+3}△=3−4+3
\mathsf{\triangle=2}△=2
\begin{gathered}⟹ \begin{gathered}\mathsf{{\triangle}_x=\left|\begin{array}{ccc}6&1&1\\12&3&1\\14&3&2\end{array}\right|}\end{gathered} \end{gathered}
⟹
△
x
=
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6
12
14
1
3
3
1
1
2
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\mathsf{{\triangle}_x=6(6-3)-1(24-14)+1(36-42)}△
x
=6(6−3)−1(24−14)+1(36−42)
\mathsf{{\triangle}_x=18-10-6}△
x
=18−10−6
\mathsf{{\triangle}_x=2}△
x
=2
\begin{gathered}\begin{gathered}\mathsf{{\triangle}_y=\left|\begin{array}{ccc}1&6&1\\3&12&1\\2&14&2\end{array}\right|}\end{gathered} \end{gathered}
△
y
=
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1
3
2
6
12
14
1
1
2
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\mathsf{{\triangle}_y=1(24-14)-6(6-2)+1(42-24)}△
y
=1(24−14)−6(6−2)+1(42−24)
\mathsf{{\triangle}_y=10-24+18}△
y
=10−24+18
\mathsf{{\triangle}_y=4}△
y
=4
\mathsf{{\triangle}_z=1(42-36)-1(42-24)+6(9-6)}△
z
=1(42−36)−1(42−24)+6(9−6)
\mathsf{{\triangle}_z=6-18+18}△
z
=6−18+18
\mathsf{{\triangle}_z=6}△
z
=6
\textsf{By Cramer's rule}By Cramer’s rule
\mathsf{x=\dfrac{\triangle_x}{\triangle}}x=
△
△
x
\mathsf{x=\dfrac{2}{2}=1}x=
2
2
=1
\mathsf{y=\dfrac{\triangle_y}{\triangle}}y=
△
△
y
→\mathsf{y=\dfrac{4}{2}=2}→y=
2
4
=2
→\mathsf{z=\dfrac{\triangle_z}{\triangle}}→z=
△
△
z
→\mathsf{z=\dfrac{6}{2}=3}→z=
2
6
=3
➪\therefore\textsf{The solution is x=1, y=2 and z=3}∴The solution is x=1, y=2 and z=3