The question is incomplete. The complete question is as follows:
Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.
![\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C-6%26-9%5Cend%7Barray%7D%5Cright%5D)
· X·![\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-3%5C%5C7%26-6%5Cend%7Barray%7D%5Cright%5D)
=<em>I</em>.
First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means
<em>I</em> = ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
So, · X· =
Isolating the X, we have
X·= -
Resolving:
X·= ![\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2-1%268-0%5C%5C-6-0%26-9-1%5Cend%7Barray%7D%5Cright%5D)
X·=![\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%268%5C%5C-6%26-10%5Cend%7Barray%7D%5Cright%5D)
Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,
X=⁻¹·
Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.
So,
·![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
=
9a - 3b = 1
7a - 6b = 0
9c - 3d = 0
7c - 6d = 1
Resolving these equations, we have a=
; b=
; c=
and d=
. Substituting:
X= ![\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B2%7D%7B11%7D%20%26%5Cfrac%7B-1%7D%7B11%7D%20%5C%5C%5Cfrac%7B7%7D%7B33%7D%26%5Cfrac%7B-3%7D%7B11%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
·
Multiplying the matrices, we have
X=![\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B8%7D%7B11%7D%20%26%5Cfrac%7B26%7D%7B11%7D%20%5C%5C%5Cfrac%7B39%7D%7B11%7D%26%5Cfrac%7B198%7D%7B11%7D%20%20%5Cend%7Barray%7D%5Cright%5D)
14% because 7/50 divided by 100 equals 14.
The correct answer is: [D]: "17" .
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The radius is: " 17" .
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Note:
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The formula/equation for the graph of a circle is:
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(x − h)² +<span> </span> (y − k)² = r² ;
in which:
" (h, k) " ; are the coordinate of the point of the center of the circle;
"r" is the length of the "radius" ; for which we want to determine;
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We are given the following equation of the graph of a particular circle:
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→ (x − 4)² + (y + 12)² = 17² ;
which is in the correct form:
→ " (x − h)² + (y − k)² = r² " ;
in which: " h = 4 " ;
" k = -12" ;
"r = 17 " ; which is the "radius" ; which is our answer.
→ { Note that: "k = NEGATIVE 12" } ;
→ Since the equation <u>for this particular circle</u> contains the expression: _________________________________________________________
→ "...(y + k)² ..." ;
[as opposed to the standard form: "...(y − k)² ..." ] ;
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→ And since the coordinates of the center of a circle are represented by:
" (h, k) " ;
→ which are: " (4, -12) " ; (<u>for this particular circle</u>) ;
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→ And since: " k = -12 " ; (<u>for this particular circle</u>) ;
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then:
" [y − k ] ² = [ y − (k) ] ² = " [ y − (-12) ] ² " ;
= " ( y + 12)² " ;
{NOTE: Since: "subtracting a negative" is the same as "adding a positive" ;
→ So; " [ y − (-12 ] " = " [ y + (⁺ 12) ] " = " (y + 12) "
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Note: The above explanation is relevant to confirm that the equation is, in fact, in "proper form"; to ensure that the: radius, "r" ; is: "17" .
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→ Since: "r = 17 " ;
→ The radius is: " 17 " ;
which is: Answer choice: [D]: "17" .
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Answer: A. sqrt of 500.
Step-by-step explanation: If you simplify the sqrt of 500, you will get 10 sqrt of 5
