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:
D. 42
Step-by-step explanation:
Base = 10+4 (ft)
10+4 = 14
Height = 3
14 x 3 = 42
Answer:
Step-by-step explanation:
step a system of two equations c = child ticket a = adult ticket
eq 1) 2c + 1a = 8.2 multiply by 2
eq 2) 3c + 2a = 14.1
I will multiply eq 1 times TWO and subtract eq 2 from eq 1a)
eq 1a) 4c + 2a = 16.4
eq 2) 3c + 2a = 14.1
subtract (4c - 3c) + (2a -2a) = 16.4 - 14.1
c + 0 = 2.3 euros for one child ticket
Now find the adult ticket price, plug 2.3 for c into eq 1)
eq 1) 2c + 1a = 8.2
eq 1) 2(2.3) + 1a = 8.2 solve for a
4.6 + a = 8.2 substract 4.6 from both sides
a = 8.2 - 4.6
= 3.6 euros for one adult ticket
double check using eq 2) we know c and a values
eq 2) 3c + 2a = 14.1
eq 2) 3(2.3) + 2(3.6) = 14.1
6.9 + 7.2 = 14.1
14.1 = 14.1
Answer:
18 1/2 miles
Step-by-step explanation:
3 7/10 * 5
Change to an improper fraction
(10* 3 + 7)/10 * 5
37/10 *5
Rewriting
37 * 5/10
37/2
Changing back to an mixed number
2 goes into 37 18 times with 1 left over
18 1/2
