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.
· X·
=<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> =
So,
· X·
= ![\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)
Isolating the X, we have
X·
=
- ![\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)
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·![\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, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.
So,
·
=![\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)
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}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)
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:
The dimensions of the yard are W=20ft and L=40ft.
Step-by-step explanation:
Let be:
W: width of the yard.
L:length.
Now, we can write the equation of that relates length and width:
(Equation #1)
The area of the yard can be expressed as (using equation #1 into #2):
(Equation #2)
Since the Area of the yard is
, then equation #2 turns into:

Now, we rearrange this equation:

We can divide the equation by 5 :

We need to find the solution for this quadratic. Let's find the factors of 160 that multiplied yields -160 and added yields -12. Let's choose -20 and 8, since
and
. The equation factorised looks like this:

Therefore the possible solutions are W=20 and W=-8. We discard W=-8 since width must be a positive number. To find the length, we substitute the value of W in equation #1:

Therefore, the dimensions of the yard are W=20ft and L=40ft.
Answer:
The Answer would be 1.4f-18
Step-by-step explanation:
-2.3f+0.9f-14-4
then you do
-1.4f-14-4
Then u would get 1.4f-18.
Answer:
x=35
Step-by-step explanation:
180-147=33 (angles in a straight line)
33+3x+42=180 (angles in a triangle)
3x+42=147
147-42=105
105/3=35
X² - 6x - 9 = 0
x² - 3x - 3x - 9 = 0
x (x-3) -3(x + 3) = 0
(x - 3)² = 0
In short, Your Answer would be Option A
Hope this helps!