Please solve the problem

Mathematics

1 answer:

jek_recluse [69]2 years ago

3 0

Treat the matrices on the right side of each equation like you would a constant.

Let 2X + Y = A and 3X - 4Y = B.

Then you can eliminate Y by taking the sum

4A + B = 4 (2X + Y) + (3X - 4Y) = 11X

==> X = (4A + B)/11

Similarly, you can eliminate X by using

-3A + 2B = -3 (2X + Y) + 2 (3X - 4Y) = -11Y

==> Y = (3A - 2B)/11

It follows that

$X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}$

Similarly, you would find

$Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}$

You can solve the second system in the same fashion. You would end up with

$P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}$

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aivan3 [116]

I think B i am not 100 sure

5 0

2 years ago

Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p

zalisa [80]

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

Statement 1: "The pen is 3 feet wider than it is long"

$Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3$

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Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"

$Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y$

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Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."

$Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\$

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Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."

$Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\$