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Talja [164]
3 years ago
7

Solve the following matrix system: The explanation how you did it would help! Thanks in advance!

Mathematics
1 answer:
mestny [16]3 years ago
7 0

Answer:

See below

Step-by-step explanation:

X - Y =  \begin{pmatrix} 1 & 2 & 7 \\  - 1 & 3 & 5 \\ 3 & 1 & 7 \end{pmatrix} \\\\ X  +  Y =  \begin{pmatrix} 1 & 0 & 1 \\  1 & 1 & 3 \\  - 1 & 3 & 3 \end{pmatrix} \\  adding \: both \: the \: matrices \\  \\ X   -   Y + X  +  Y \\  = \begin{pmatrix} 1 & 2 & 7 \\  - 1 & 3 & 5 \\ 3 & 1 & 7 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 1 \\  1 & 1 & 3 \\  - 1 & 3 & 3 \end{pmatrix} \\\\  2X= \begin{pmatrix} 1 + 1 & 2 + 0 & 7 + 1 \\  - 1 + 1 & 3 + 1 & 5  + 3\\ 3 - 1 & 1 + 3 & 7 + 3 \end{pmatrix} \\  \\ 2X= \begin{pmatrix} 2 & 2 & 8 \\  0 & 4 & 8\\ 2 & 4 & 10 \end{pmatrix} \\  \\ X=  \frac{1}{2} \begin{pmatrix} 2 & 2 & 8 \\  0 & 4 & 8\\ 2 & 4 & 10 \end{pmatrix}\\  \\ X=   \huge\begin{pmatrix} \frac{2}{2} & \frac{2}{2} & \frac{8}{2} \\  \\  \frac{0}{2} & \frac{4}{2} & \frac{8}{2}\\ \\  \frac{2}{2} & \frac{4}{2} & \frac{10}{2} \end{pmatrix} \\  \\  \huge  \red{X=   \begin{pmatrix} 1 & 1 & 4\\  \\ 0 & 2 & 4\\ \\  1 & 2& 5 \end{pmatrix}} \\  \\ subtracting \: the \: value \: of \: x \: from \: matrix \: (2) \\  \\ X  +  Y  - X \\ =  \begin{pmatrix} 1 & 0 & 1 \\  1 & 1 & 3 \\  - 1 & 3 & 3 \end{pmatrix} - \begin{pmatrix} 1 & 1 & 4 \\ 0 & 2 & 4\\  1 & 2& 5 \end{pmatrix} \\  \\ Y = \begin{pmatrix} 1 - 1 & 0 - 1 & 1 - 4 \\  1  - 0& 1 - 2 & 3 - 4 \\  - 1 - 1 & 3 - 2 & 3 - 5 \end{pmatrix} \\  \\  \huge  \purple{Y = \begin{pmatrix} 0 & - 1 &  - 3 \\  1  &  - 1  &  - 1 \\  - 2 & 1 &  - 2 \end{pmatrix}}

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brilliants [131]
\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ -9}}\quad ,&{{ -3}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies -6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-3)=-6[x-(-9)]
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8 0
3 years ago
A system for tracking ships indicated that a ship lies on a hyperbolic path described by 5x2 - y2 = 20. the process is repeated
zysi [14]
Answer:
The ship is located at (3,5)

Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III

Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II

To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.

Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9

We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3

Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5

Based on the above, the position of the ship is (3,5).

Hope this helps :)
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3 years ago
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Answer:

Step-by-step explanation:

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link to learning

4 0
2 years ago
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Ber [7]

Answer:

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same steps for g(x)=-2x-4

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5 0
3 years ago
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Answer:

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