1
2
3x
1 7
X +
5 3
+
+
4.
+
2x2
-X + 5
X
3
6
2.
Answers
Answer:
your name is shree and my name is shree
Answer:
3x1 + 2x2 -x3= -15
5x1+ 3x2 + 2x3 =0
3x1+ x2 +3x3=11
11x1+7x2 = -30
\begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 5 & 3 & 2 \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\ 3 & 1 & 3 \ \ \ \ \ \ \ \ \ \ \ \ 11 \\ 11 & 7 & 0 \ \ \ \ \ \ \ -30 \end{bmatrix} =
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⎢
⎢
⎢
⎡
3
5
3
11
2
3
1
7
−1 −15
2 0
3 11
0 −30
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⎤
= \begin {Vmatrix} R_2 =5R_1-3R_2 \\ R_3 = R_1-R_3 \\ R_4 = 5R_4-4R_2 \end{Vmatrix} =
∥
∥
∥
∥
∥
∥
∥
R
2
=5R
1
−3R
2
R
3
=R
1
−R
3
R
4
=5R
4
−4R
2
∥
∥
∥
∥
∥
∥
∥
= \begin {bmatrix} 3 & 2 & -1 \ \ \ \ \ \ -15 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 1 & -2 \ \ \ \ \ \ \ \ \ \ \ 26 \\ 0 & 2 & -22 \ \ \ \ \ \ \ -150 \end{bmatrix}=
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⎢
⎢
⎢
⎡
3
0
0
0
2
1
1
2
−1 −15
−1 −75
−2 26
−22 −150
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⎥
⎥
⎥
⎤
=
\begin {Vmatrix} R_1 =R_1-2R_2 \\ R_2 =R_2-R_3 \\ R_3 = 2R_3-R_4 \\ R_4 = 2R_2- R_4 \end{Vmatrix} =
∥
∥
∥
∥
∥
∥
∥
∥
∥
R
1
=R
1
−2R
2
R
2
=R
2
−R
3
R
3
=2R
3
−R
4
R
4
=2R
2
−R
4
∥
∥
∥
∥
∥
∥
∥
∥
∥
= \begin {bmatrix} 3 & 0 & 11 \ \ \ \ \ \ -135 \\ 0 & 1 & -1 \ \ \ \ \ \ \ -75\\ 0 & 0 & 18 \ \ \ \ \ \ \ \ \ \ \ 202 \\ 0 & 0 & 20 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{bmatrix}=
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⎢
⎢
⎢
⎡
3
0
0
0
0
1
0
0
11 −135
−1 −75
18 202
20 0
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⎥
⎥
⎥
⎤
= \begin {Vmatrix} R_1 =\frac{R_1}{3} \\ R_2 =R_2-R_3 \\ R_3 = \frac{R_3}{18} \\ R_4 = 10R_3-9R_4 \end{Vmatrix} =
∥
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∥
∥
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∥
R
1
=
3
R
1
R
2
=R
2
−R
3
R
3
=
18
R
3
R
4
=10R
3
−9R
4
∥
∥
∥
∥
∥
∥
∥
∥
∥
=
\begin {bmatrix} 1 & 0 & \frac{11}{3} \ \ \ \ \ \ -\frac{135}{3} \\ 0 & 1 & -1 \ \ \ \ \ \ -75\\ 0 & 0 & 1 \ \ \ \ \ \ \ \ \ \ \frac{101}{9} \\ 0 & 0 & 0 \ \ \ \ \ \ \ \ \ 2020 \end{bmatrix};
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⎢
⎢
⎢
⎡
1
0
0
0
0
1
0
0
3
11
−
3
135
−1 −75
1
9
101
0 2020
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⎥
⎥
⎥
⎤
;
In R_4R
4
0 \ne 20200
=2020 . So this system has no solution.