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)
Answer:
(2,4)
Step-by-step explanation:
When graphed the lines intersect at point (2,4) which is the solution.
9514 1404 393
Answer:
C) 12y 8m
Step-by-step explanation:
The amount of principal P at compound monthly at interest rate r per year is given by ...
A = P(1 +r/12)^(12t) . . . . after t years
Here, we want to find t, so ...
A/P = (1 +r/12)^(12t)
log(A/P) = (12t)·log(1 +r/12)
t = log(A/P)/(12·log(1 +r/12))
Filling in the given values, we find t to be ...
t = log(8000/4000)/(12·log(1 +0.055/12)) ≈ 12.6315 ≈ 12 years 7.6 months
It will take about 12 years 8 months to double the money.
Collect like terms and you'll get -2 - 3(a - b)
Answer:
Root 181
Step-by-step explanation:
Using distance formula, we know that the answer will be the square root of 181. Unfortunately, this is not a perfect square, and it is prime, so that number will remain the same.
Distance Formula: 
Hope this helped!
