# Linear Equations

• Sep 13th 2007, 04:20 PM
Thomas
Linear Equations
http://tunerspec.ca/school/math.jpg

For the sixth question, I got:
$cos a = \frac{-3}{8}$
$cos b = \frac{-5}{4}$
$sin 2a = \frac{7}{8}$

Whenever I use the answer I got, it says it's incorrect, but every time I do it I get that answer.
• Sep 13th 2007, 08:49 PM
Soroban
Hello, Thomas!

Quote:

6. Solve the following system.

. . $\begin{array}{ccccccccc}
3\cos a & + & \cos b & + & 5\sin2a & = & 2 & & \\
& & & & \cos a & = & \cos b & + & \sin2a \\
& & 6\cos a & + & 13\sin2a & = & \text{-}7\cos b & & \end{array}$

We have: . $\begin{array}{ccccccc}
3\cos a & + & \cos b & + & 5\sin2a & = & 2 \\
\cos a & - & \cos b & - &\sin2a & = & 0 \\
7\cos a & + & 7\cos b & + & 13\sin2a & = & 0 \end{array}$

And then: . $\begin{vmatrix}
\;3 & 1 & 5 & | & 2 \;\\ \;1 & \text{-}1 & \text{-}1 & | & 0 \;\\ \;7 & 7 & 13 & | & 0\;\end{vmatrix}$

$\begin{array}{c}\text{Switch}\\ R_1\text{ and }R_2 \\ \\ \end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \;\\\;3 & 1 & 5 & | & 2 \;\\\;7 & 7 & 13 & | & 0\;\end{vmatrix}$

$\begin{array}{c} \\ R_2 - 3R_1 \\ R_3 - 7R_1\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0\; \\\;0 & 4 & 8 & | & 2 \;\\\;0 & 14 & 20 & | & 0\;\end{vmatrix}$

. . . $\begin{array}{c} \\ \frac{1}{4}R_2 \\ \frac{1}{2}R_3\end{array}\;\begin{vmatrix}\;1 & \text{-}1 & \text{-}1 & | & 0 \\0 & 1 & 2 & | & \frac{1}{2} \\0 & 7 & 10 & | & 0\end{vmatrix}$

$\begin{array}{c}R_1+R_2 \\ \\ R_3 - 7R_2\end{array}\;\begin{vmatrix}\;1 & 0 & 1 & | & \frac{1}{2}\;\\
\;0 & 1 & 2 & | & \frac{1}{2}\; \\
\;0 & 0 & \text{-}4 & | & \text{-}\frac{7}{2}\; \end{vmatrix}$

. . . $\begin{array}{c} \\ \\ \text{-}\frac{1}{4}R_3\end{array}\;
\begin{vmatrix}1 & 0 & 1 & | & \frac{1}{2}\\
0 & 1 & 2 & | & \frac{1}{2}\\
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}$

$\begin{array}{c}R_1-R_3 \\ R_2 - 2R_3 \\ \\ \end{array}\;\begin{vmatrix}1 & 0 & 0 & | & \text{-}\frac{3}{8} \\
0 & 1 & 0 & | & \text{-}\frac{5}{4} \\
0 & 0 & 1 & | & \frac{7}{8}\end{vmatrix}$

Therefore: . $\begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} \\ \\\cos b & = & \text{-}\frac{5}{4} \\ \\ \sin2a & = & \frac{7}{8} \end{Bmatrix}$

• Sep 14th 2007, 12:45 PM
Thomas
Thank you for the help!

Is anyone able to help me on the other ones?
• Sep 15th 2007, 02:52 PM
Thomas
Bump, seems that ISN'T the answer. Anyone have any other suggestions?
• Sep 15th 2007, 03:32 PM
Soroban

Quote:

Therefore: . $\begin{Bmatrix}\cos a & = & \text{-}\frac{3}{8} & [1] \\ \\\cos b & = & \text{-}\frac{5}{4} & [2]\\ \\ \sin2a & = & \frac{7}{8} & [3]\end{Bmatrix}$

These answers check out perfectly in the three equations.

If there are objections, they should be directed to the problem itself.
Either there are typos in the original statement
. . or the problem was deliberately created to cause confusion.

If [1] is true, then [3] cannot be true.

Also, [2] has no solution in real numbers.

The system is inconsistent and has complex roots.
. . There are no correct answers.

• Sep 15th 2007, 03:53 PM
Thomas
I also agree with your work, I just don't get how we're supposed to answer this question as they are asking for a number (answers are submitted online.) I inputted -3/8 and it said I had the wrong answer.