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iris [78.8K]
3 years ago
15

How to solve the matrix equation for matrix Y.simplify all elements

Mathematics
1 answer:
marta [7]3 years ago
4 0

Given <em>YA</em> = <em>B</em>, you can solve for <em>Y</em> by multiplying by <em>A </em>⁻¹ on the right (on both sides of the equation). So we have

<em>YA</em> = <em>B</em>   ==>   (<em>YA</em>) <em>A </em>⁻¹ = <em>BA </em>⁻¹   ==>   <em>Y</em> (<em>AA </em>⁻¹) = <em>BA </em>⁻¹   ==>   <em>Y</em> = <em>BA </em>⁻¹

provided that the inverse of <em>A</em> exists. In this case, det(<em>A</em>) = 5 ≠ 0, so the inverse does exist, and

A=\begin{bmatrix}-1&-4\\0&-5\end{bmatrix} \implies A^{-1}=\dfrac1{\det(A)}\begin{bmatrix}-5&0\\4&-1\end{bmatrix} = \begin{bmatrix}-1&0\\\frac45&-\frac15\end{bmatrix}

Then

Y=\begin{bmatrix}5&-5\\8&-8\end{bmatrix}A^{-1} = \begin{bmatrix}-5&5\\-8&\frac{24}5\end{bmatrix}

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Answer:

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Step-by-step explanation:

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What is the zero of the function d=70-2.4t
AVprozaik [17]

Answer:  The "zero" is:  " 2 \frac{11}{12} " ;  or, write as:  " 2.92 " .

_____________________________________

              →    " t = 2 \frac{11}{12} " ;  or, write as:  " t = 2.92 " .

_____________________________________

Step-by-step explanation:

_____________________________________

Letting assume that this given function is supposed to be written as:

 "distance as a function of time" ;  that is:

           d(t) = 70 - (2.4)t ;

  →  since distance, "d" is the dependent variable (cannot be "manipulated or controlled") and as such, belongs on the "y-axis"—as the "dependent variable" ;  whereas as time, "t" ; can be somewhat controlled (with respect to distance, at list as a starting point); and as such belongs on the "x-axis" as the "independent variable" .

Since no "specific units" are given to us in the problem for Either "distance" or "time" ; we shall use the term "units" to describe their values.

We have:

 d(t) = 70 - (2.4)t ;

Let us rearrange this:

70 - (2.4)t  ;   ↔

  =  70  +  (- 2.4)t  ;  ↔

  =  (-2.4)t  +  70 ;

And rewrite the function:

          →   d(t) = (-2.4)t + 70 ;

To find the "zero" ; or "zeros" ; of this function; set "d(t)" equal to "zero" ; that is; "0" ; and solve for  the value(s) for "t" when "d(t)" = 0 :

          →  0 = -2.4(t) + 70 ;  ↔

   Rewrite as:

          →  -2.4(t) + 70 = 0 ;  

For simplicity;  let us multiply Each side of the equation by "10" ;

to get rid of the decimal value:

    10*[ (-2.4)t) + 70 ] = 10 * [0] ;

From the left-hand side of the question:

Note the "distributive property" of multiplication:

    a(b + c) = ab + ac ;

As such:

   10* [-2.4(t) + 70 ] =

    [10* -2.4(t)] + [10 * 70] =

      -24t + 70 ;

From the "right-hand side" of the equation:

    10 * 0 = 0 .

__________________________________

So; we rewrite the equation as:

  -24t + 70 = 0 ;

__________________________________

Solve for " t " ;  

 -24t + 70 = 0 ;

Subtract "70" from Each Side of the equation;

  -24t + 70 - 70 = 0 - 70 ;  

to get:

  -24t = -70 ;

Now, let's multiply each side of the equation by "-1" ;

       to get rid of the "negative values" ;

  -1* (-24t) = -1(-70) ;

to get:

   24t = 70 ;

Now, let's divide Each Side of the equation by "24" ;

to isolate:  "t" ;  on one side of the equation; & to solve for "t" ;

   24t / 24 = 70/24 ;

to get:  

         t = 70/24 ;

To simplify:  either:  

1)  use calculator:  70 ÷ 24 = 2.916666666 ;

                             →  round to:  2.92 ;

                             →  t ≈ 2.92 ;

 or:  " \frac{70}{24} =\frac{(70/2)}{(24/2)}=\frac{35}{12} ;

→  write as simplified improper fraction:  " t = \frac{35}{12} "

→  or:  write as mixed number:

           →  " \frac{35}{12} = 35 ÷ 12 =

                              2 R 11

                     12 ⟌35

                          <u>- 24  </u>

                             1 1

         

           →  write as:  " 2\frac{11}{12} " ;

                                                               →  " t = 2 \frac{11}{12} " .

_____________________________________

Hope this is helpful to you.

        Wishing you the best!

_____________________________________

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Which triangle results from a reflection across the line x = 1?
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Answer:

Correct answer is option D.

Step-by-step explanation:

Given that \triangle ABC in the image 1 attached.

If we have a look at the image attached, the coordinates are:

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<em>To find reflection</em> of a point across any line, the distance of points from the line must be same.

Point A(1,1) lies on the line x = 1, so its reflection A' will be at the same point A'(1,1).

Point C(2,5) is at a distance 1 from x = 1 on right side, so C' will be 1 distance on the left side of x = 1 i.e. 1 will be subtracted from its x coordinate.

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