# Solve for A. $\Bigl[\begin{smallmatrix}9&9\\-9&0\end{smallmatrix}\Bigr]=4A-\Bigl[\begin{smallmatrix}2&-2\\0&2\end{smallmatrix}\Bigr]A$ [closed]

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?

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

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• Have you tired to do it in components? – Jack Feb 23 '17 at 15:55
• It's 4 multiplied by the identity matrix x A – naveen dankal Feb 23 '17 at 15:56
• It works by the definition of scalar multiplication of matrices. – Jack Feb 23 '17 at 16:08
• 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 $is the the matrix wit 4 in its diagonal. – fleablood Feb 23 '17 at 16:24 ## 2 Answers Note that$A=IA$and$kI = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$for all$k$, so it follows that $$4A = 4IA = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}A$$ • 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. – Jack Feb 23 '17 at 16:25 • @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. – Clive Newstead Feb 23 '17 at 16:57 • 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). – Jack Feb 23 '17 at 17:19 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.