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So I was just studying how to solve this problem in linear algebra but I am very confused how the user got from the first line to the second. Shouldn't there be matrix multiplication for those to matrices or not? I'm just confused what operation is being performed.

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2 Answers 2

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Yes! That is indeed matrix multiplication. Recall that the $(i,j)^{th}$ position in the resultant matrix is dot product of the vectors formed by the $i^{th}$ row of the left matrix and the $j^{th}$ column of the right matrix.

If we let the resultant matrix be called $C$, we get that

$$C_{1,1}=\langle 1,2\rangle\cdot\langle 1,-3\rangle=1-6 \;(=-5)\\ C_{1,2}=\langle 1,2\rangle\cdot\langle 2,1\rangle=2+2 \;(=4)\\ C_{2,1}=\langle 3,-4\rangle\cdot\langle 1,-3\rangle=3+12 \;(=15)\\ C_{2,2}=\langle 3,-4\rangle\cdot\langle 2,1\rangle=6-4 \;(=2)$$

There are other ways to think of matrix multiplication, but I think this method is most clear. Please let me know if you are confused.

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Yes, the matrices are being multiplied. Remember that the product of matrices is given by:

$$\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\left( \begin{array}{cc} e & f \\ g & h \end{array} \right)= \left( \begin{array}{cc} ae+bg & af+bh \\ ce+dg & cf+dh \end{array} \right) $$

Now in your case: $$\left( \begin{array}{cc} 1 & 2 \\ 3 & -4 \end{array} \right)\left( \begin{array}{cc} 1 & 2 \\ -3 & 1 \end{array} \right)= \left( \begin{array}{cc} 1+(-6) & 2+2 \\ 3+12 & 6+(-4) \end{array} \right)= \left( \begin{array}{cc} -5 & 4 \\ 15 & 2 \end{array} \right) $$

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