The question is missing parts. Here is the complete question.
Let M = ![\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
. Find 
and 
such that 
, where 
is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.
Answer: 
Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:
![\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
![M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D)
![M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]](https://tex.z-dn.net/?f=M%5E%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D)
Solving equation:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2Bc_%7B1%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%265%5C%5C-1%26-4%5Cend%7Barray%7D%5Cright%5D%20%2Bc_%7B2%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.
So, the equation is:
![\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D31%2610%5C%5C-2%2615%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6c_%7B1%7D%265c_%7B1%7D%5C%5C-1c_%7B1%7D%26-4c_%7B1%7D%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dc_%7B2%7D%260%5C%5C0%26c_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
And the system of equations is:
There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:
With , substitute in one of the equations and find :
<u>For the equation, </u><u> and </u><u />
Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least...
<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>
<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>
<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>
<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>
<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>
<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>
<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>
<span>Now apply the Law of Sines to find the other two angles. </span>
<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>
<span>Two down, one to go... </span>
<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>
<span>So your three angles are: </span>
<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
Answer: 18
Step-by-step explanation: well 9 x 12 = 108 then divide it by 6 and yyou get 18 wich is 6 x 3
Answer:
The correct answer is Never
Answer:
Yes he have enough white paint.
Step-by-step explanation:
We are given
6 gallons of light green paint use → 
gallons of white paint
using the unitary method
1 gallons of white paint → 6*8/3 gallons of light green paint
→ 16 gallons of light green paint
but Liam have 9/16 gallons of white paint
9/16 gallons of white paint → 16* 9/16 gallons of light green paint
→ 9 gallons of light green paint
Yes Liam have enough white paint to make 8 gallons of light green paint.
