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I didn't know how to evaluate $$4A-\begin{bmatrix} 2 &-2 \\ 0 & 2 \end{bmatrix}A$$

and so I looked in the solutions, and what they did was they rewrote $4A$ as

$$\begin{bmatrix} 4 &0 \\ 0 & 4 \end{bmatrix}A-\begin{bmatrix} 2 &-2 \\ 0 & 2 \end{bmatrix}A$$

Is this allowed? Why does this work?

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closed as off-topic by Jack, Daniel W. Farlow, Namaste, Leucippus, C. Falcon Feb 24 '17 at 0:10

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  • $\begingroup$ Have you tired to do it in components? $\endgroup$ – Jack Feb 23 '17 at 15:55
  • $\begingroup$ It's 4 multiplied by the identity matrix x A $\endgroup$ – naveen dankal Feb 23 '17 at 15:56
  • $\begingroup$ It works by the definition of scalar multiplication of matrices. $\endgroup$ – Jack Feb 23 '17 at 16:08
  • $\begingroup$ Assuming A is a 2x2 matrix (which it must be for the expression to make sense) then 4A = 4(I$\times$ A)= (4I)A$ and $4I $ is the the matrix wit 4 in its diagonal. $\endgroup$ – fleablood Feb 23 '17 at 16:24
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Note that $A=IA$ and $kI = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$ for all $k$, so it follows that $$4A = 4IA = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}A$$

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  • $\begingroup$ The identity $4(IA)=(4I)A$ is implicitly used, which is not obvious and one might eventually end up with proving $4A=4(IA)$ by definition. $\endgroup$ – Jack Feb 23 '17 at 16:25
  • $\begingroup$ @Jack: I don't see the problem. Obvious or not, when doing matrix algebra (like in this question), it is fine to assume basic rules of matrix multiplication and scalar multiplication without re-proving them every time. $\endgroup$ – Clive Newstead Feb 23 '17 at 16:57
  • $\begingroup$ Surely my comment was not an objection to your answer. "it is fine to assume basic rules of matrix multiplication and scalar multiplication without re-proving them every time" I would assume that this is not a suggestion for one who does not know the "basic rules" or does not know the necessity of these rules (yet). $\endgroup$ – Jack Feb 23 '17 at 17:19
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It follows from the definition of scalar multiplication of matrices:

$$ k\begin{pmatrix} a&b\\ c&d \end{pmatrix}= \begin{pmatrix} ka&kb\\ kc&kd \end{pmatrix}. $$ By this definition, for any $2\times 2$ matrix $A$ and any scalar $k$, one can check by direct calculation that $$ kA=(kI)A. $$ Of course one can write $$ kA=k(IA)=(kI)A, $$ if one knows

  • how $kI$ is defined
  • why one has "associativity" in the second identity.
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