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3 years ago

Solve the equation for X.

Mathematics

1 answer:

OlgaM077 [116]3 years ago

7 0

Answer:

$\large\boxed{\left[\begin{array}{ccc}10&-30\\-10&-2\end{array}\right]}$

Step-by-step explanation:

$X-3\left[\begin{array}{ccc}2&-8\\-4&2\end{array}\right] =\left[\begin{array}{ccc}4&-6\\2&-8\end{array}\right]\\\\X-\left[\begin{array}{ccc}(3)(2)&(3)(-8)\\(3)(-4)&(3)(2)\end{array}\right]=\left[\begin{array}{ccc}4&-6\\2&-8\end{array}\right]\\\\X-\left[\begin{array}{ccc}6&-24\\-12&6\end{array}\right]=\left[\begin{array}{ccc}4&-6\\2&-8\end{array}\right]$

$X-\left[\begin{array}{ccc}6&-24\\-12&6\end{array}\right]+\left[\begin{array}{ccc}6&-24\\-12&6\end{array}\right]=\left[\begin{array}{ccc}4&-6\\2&-8\end{array}\right]+\left[\begin{array}{ccc}6&-24\\-12&6\end{array}\right]\\\\X=\left[\begin{array}{ccc}4+6&(-6)+(-24)\\2+(-12)&-8+6\end{array}\right]\\\\X=\left[\begin{array}{ccc}10&-30\\-10&-2\end{array}\right]$

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Read 2 more answers

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DaniilM [7]

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Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .

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