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I'm currently applying for a PHD and may need to take a further maths course along with it and so I'm practicing in advance so that I can be ahead of myself when the time comes.

I've been working with matrices recently and have been doing well with understanding. I've just come across this question:

$$A = \begin{pmatrix} 4 & 1 \\ -3 & 5 \end{pmatrix}$$ $$C = \begin{pmatrix} 11 & -13 \\ 3 & 1 \end{pmatrix}$$

For this question (right at the back of the textbook so hard difficulty) I need to find a 2x2 matrix $X$ such that $2A - 3X = C$.

Now normally I never come to stackexchange until I have worked out an answer and am just needing a second opinion. I haven't in this situation come across a question like this however, and it's driving me loopy.

I think I'm on to the right path in terms of figuring determinants and links. What would be first step to fulfilling the criteria here? Or the most sensible step towards an answer? I'd love some guidance. Thanks!

Background - I'm a chemical engineering major but have never worked with matrices before.

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    $\begingroup$ As an aside, you will notice that your matrices appeared as (4 1)(-3 5) all in one row. Visiting this page will show you how to type mathematics properly on this site using MathJax and $\LaTeX$, to instead have your matrix appear more like how you intend, such as $\left[\begin{smallmatrix} 4&1\\-3&5\end{smallmatrix}\right]$ $\endgroup$
    – JMoravitz
    Mar 21 '16 at 17:37
  • $\begingroup$ Thanks so much for the link JMoravitz. I'm going to learn it all because this website will prove invaluable to me during me PHD. Many thanks $\endgroup$ Mar 21 '16 at 17:39
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    $\begingroup$ As for your question, you ask for a matrix such that $2A-3X=C$. Matrix equations can be manipulated in many of the same ways that polynomials of real numbers can. Rearranging gives us $2A-C=3X$ and dividing by three gives $\frac{1}{3}(2A-C)=X$. There is no need to invoke determinants or other things beyond scalar multiplication, addition, and subtraction of matrices. $\endgroup$
    – JMoravitz
    Mar 21 '16 at 17:39
  • $\begingroup$ It is worth mentioning as well that typing papers and textbooks in $\LaTeX$ is usually standard in fields which rely heavily on mathematics and equations. I personally use the packages available on miktex.org for writing up all of my homework and papers. $\endgroup$
    – JMoravitz
    Mar 21 '16 at 17:42
  • $\begingroup$ Thanks for this. I can imagine the applications it must have! I'm going to learn to become competent in its use. Bookmarked. $\endgroup$ Mar 21 '16 at 18:29
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Matrix equations can be manipulated in similar ways as polynomials of real numbers.

$2A-3X=C$ can be rearranged to

$\frac{1}{3}(2A-C)=X$

Inserting your specific matrices $A$ and $C$ and applying arithmetic will complete the question.

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    $\begingroup$ Thanks for your help here, I really appreciate it. $\endgroup$ Mar 21 '16 at 18:28
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The nice thing about matrices is that for addition, scaler multiplication, and subtraction, you can go unit by unit. For the equation $2A-3X=C$, you know that:

$$2\begin{bmatrix}4&1\\-3&5\end{bmatrix}-3\begin{bmatrix}x_1&x_2\\x_3&x_4\end{bmatrix}=\begin{bmatrix}11&-13\\3&1\end{bmatrix}$$

Hence,

$$2(4)-3(x_1)=11$$ $$2(1)-3(x_2)=-13$$ $$2(-3)-3(x_3)=3$$ $$2(5)-3(x_4)=1$$

So, all you have to do is solve for each variable, and there's your matrix.

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  • $\begingroup$ Thanks for this clear explanation, I really appreciate it. $\endgroup$ Mar 21 '16 at 18:23
  • $\begingroup$ Thanks for this helpful explanation. So if I do 8 − 3(x1) = 11 obviously I don't have a value for x so it just becomes 8-3x1 = 11 The work I've been looking at lately has looked at dividing by inverses but clearly these are not here. Does the 2x2 new matrix absolutely have to have integers as its values? $\endgroup$ Mar 21 '16 at 18:28
  • $\begingroup$ Nope, there's no need for it to have integer values, but in this case it should be. Just solve for $x_1$: $$8-3x_1=11$$ $$8=11+3x_1$$ $$-3=3x_1$$ $$x_1=-1$$ $\endgroup$
    – LuuBluum
    Mar 21 '16 at 18:29
  • $\begingroup$ Thanks for your help, I was tantalizingly close! So my new matrix values for the 2x2 are: -1, 5, -3 and 3 Again, I really appreciate the guidance $\endgroup$ Mar 21 '16 at 18:35
  • $\begingroup$ Yep, that's the solution. Things do get a bit more complicated when you start dealing with more complex matrix operations (matrix multiplication, for instance), but so long as you're dealing with basic arithmetic operators such as scaler multiplication, addition, and subtraction, going term-wise is your safest bet. $\endgroup$
    – LuuBluum
    Mar 21 '16 at 18:38

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