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)
To answer this item, we assume that the topic is in similar polygons such that the ratio of the corresponding sides should be equal. In this item,
BY/YC = AX/XC
Substituting the known values,
6/10 = 18/XC
The value of XC from the equation is 30. The answer is letter C.
Answer: b≥-3
Step-by-step explanation:
at least means that it's great or equal than -3.
b≥-3
Answer: a the first answer is right
Let 11x + 15y + 23 = 0 be equation (1)
And 7x - 2y - 20 = 0 be equation (2)
Multiply equation (1) by 2:
22x + 30y + 46 = 0
Multiply equation (2) by 15:
105x - 30y - 300 = 0
Add equations (1) and (2):
22x + 105x + 30y - 30y + 46 - 300 = 0
127x - 254 = 0
127x = 254
x = 254/127
[x = 2]
Substitute x = 2 in equation (1) to find y:
11(2) + 15y + 23 = 0
22 + 15y + 23 = 0
15y + 45 = 0
15y = -45
y = -45/15
y = -3
Therefore, x = 2 and y = -3.
