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inn [45]
2 years ago
5

Use Gauss-Jordan elimination to nd the general solution for the following system of linear equations: z2 + 3z3 ???? z4 = 0 ????z

1 ???? z2 ???? z3 + z4 = 0 ????2z1 ???? 4z2 + 4z3 ???? 2z4 = 0 (b) Give an example of a non-zero solution to the previous system of linear equations. (c) The points (1; 0; 3), (1; 1; 1), and (????2;????1; 2) lie on a unique plane a1x1 + a2x2 + a3x3 = b. Using your previous answers, nd an equation for this plane. (Hint: think about the relationship between the previous system and the one you would need to solve in this question.)

Mathematics
1 answer:
nirvana33 [79]2 years ago
6 0

Answer:

Step-by-step explanation:

Solution is attached below

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If f(x) = -7x + 5x2 + 10, what does f(-2) equal?
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Answer:

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

y=0,\:x=6

Step-by-step explanation:

Considering the system of the equations

\begin{bmatrix}2x-7y=12\\ -x+15y=-6\end{bmatrix}

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2x=12+7y

Divide both sides by 2

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\mathrm{Subsititute\:}x=\frac{12+7y}{2}

\begin{bmatrix}-\frac{12+7y}{2}+15y=-6\end{bmatrix}

-\frac{12+7y}{2}+15y=-6

\mathrm{Multiply\:both\:sides\:by\:}2

-\frac{12+7y}{2}\cdot \:2+15y\cdot \:2=-6\cdot \:2

-\left(12+7y\right)+30y=-12

-12+23y=-12

-12+23y+12=-12+12

23y=0

\mathrm{Divide\:both\:sides\:by\:}23

\frac{23y}{23}=\frac{0}{23}

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\mathrm{For\:}x=\frac{12+7y}{2}

\mathrm{Subsititute\:}y=0

x=\frac{12+7\cdot \:0}{2}

x=\frac{12}{2}

x=6

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

y=0,\:x=6

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Step-by-step explanation:

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