Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:

$ \begin{align} \begin{bmatrix} a-b & b+c\\ 3d+c & 2a-4d \end{bmatrix} \end{align} = $ $ \begin{align} \begin{bmatrix} 8 & 1\\ 7 & 6 \end{bmatrix} \end{align} $

What I have done so far is:

$ \begin{align} \begin{bmatrix} a-b-8 & b+c-1\\ 3d+c-7 & 2a-4d-6 \end{bmatrix} \end{align} = $ $ \begin{align} \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} \end{align} $


And the solving for variables,

$$ a-b-8 = 0 $$ $$ a-b = 8 $$ $$ a = \frac{8}{-b} $$


$$ b+c-1=0 $$ $$ b+c=1 $$ $$ b=\frac{1}{c} $$


$$ 3d+c-7=0 $$ $$ 3d+c=7 $$ $$ 3d=\frac{7}{c} $$ $$ d=\frac{7}{3c} $$


$$ 2a-4d-6=0 $$ $$ 2a-4d=6 $$ $$ \frac{16}{-2b}-\frac{28}{12c}=6 $$


Am I going about this correctly? Or am I just doing this completely incorrect?

share|improve this question
    
Whoa. Yes, quite a few very basic mistakes here. In fact, every single one of your subproblems is wrong; you are dividing when you should be subtracting. I hope you will not take offense but I would strongly recommend a conference with your instructor. –  Michael Grant May 22 '13 at 2:48
2  
+1 for showing your work. –  response May 22 '13 at 2:51
    
@MichaelC.Grant No none taken, we haven't done anything like this in our labs or during the lectures, and I was just guessing as to what I should be doing. –  user1327636 May 22 '13 at 2:55
    
Cool. It does seem rather unfortunate that they would foist this multiequation problem on you before you had mastered isolating variables in single equations. –  Michael Grant May 22 '13 at 2:59
    
I'm taking this course during our Spring semester, and it's normally a 4 month long class, however during the Spring it's crammed down into only 2 months. So this is most likely why we are skipping some of these types of questions. Thank you for your feedback though, I appreciate it. –  user1327636 May 22 '13 at 3:01

2 Answers 2

up vote 3 down vote accepted

Error 1: $a-b=8$ is not the same as $a = \frac{8}{-b}$ but $a = 8 + b$

Error 2: $b+c=1$ is not the same as $b = \frac{1}{c}$ but $b = 1-c$

Error 3: $3d+c=7$ ... same as errors 1 and 2

Error 4: $2a-4d=6$ Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error

share|improve this answer

How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. That is, rewrite the system as follows:

$ \begin{align} \begin{bmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 3 \\ 2 & 0 & 0 & -4 \\ \end{bmatrix} \end{align} $ $ \begin{align} \begin{bmatrix} a \\ b \\ c \\ d \\ \end{bmatrix} \end{align} = $ $ \begin{align} \begin{bmatrix} 8 \\ 1\\ 7 \\ 6 \end{bmatrix} \end{align} $

And then solve.

share|improve this answer
1  
Thanks for replying, I didn't even think of trying to solve it this way. I'll keep it in mind for the rest of my exercises. –  user1327636 May 22 '13 at 3:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.