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)
Step-by-step explanation:
The equation of a circle can be the expanded form of
\large \text{$(x-a)^2+(y-b)^2=r^2$}(x−a)
2
+(y−b)
2
=r
2
where rr is the radius of the circle, (a,\ b)(a, b) is the center of the circle, and (x,\ y)(x, y) is a point on the circle.
Here, the equation of the circle is,
\begin{gathered}\begin{aligned}&x^2+y^2+10x-4y-20&=&\ \ 0\\ \\ \Longrightarrow\ \ &x^2+y^2+10x-4y+25+4-49&=&\ \ 0\\ \\ \Longrightarrow\ \ &x^2+y^2+10x-4y+25+4&=&\ \ 49\\ \\ \Longrightarrow\ \ &x^2+10x+25+y^2-4y+4&=&\ \ 49\\ \\ \Longrightarrow\ \ &(x+5)^2+(y-2)^2&=&\ \ 7^2\end{aligned}\end{gathered}
⟹
⟹
⟹
⟹
x
2
+y
2
+10x−4y−20
x
2
+y
2
+10x−4y+25+4−49
x
2
+y
2
+10x−4y+25+4
x
2
+10x+25+y
2
−4y+4
(x+5)
2
+(y−2)
2
=
=
=
=
=
0
0
49
49
7
2
From this, we get two things:
\begin{gathered}\begin{aligned}1.&\ \ \textsf{Center of the circle is $(-5,\ 2)$.}\\ \\ 2.&\ \ \textsf{Radius of the circle is $\bold{7}$ units. }\end{aligned}\end{gathered}
1.
2.
Center of the circle is (−5, 2).
Radius of the circle is 7 units.
Hence the radius is 7 units.
Answer:
14
Step-by-step explanation:
