# Collecting Matrices when solving laplace transform

\begin{align} \tag{1} \dot x&=Ax\\ \tag{2} sX(s)-x(0)&=AX(s) \\ \tag{3} (sI-A)X(s)&=x(0) \end{align} Considering these equations, how can we go from $(2)$ to $(3)$?

My question is why was the identity matrix $I$ introduced here? My concern is about introducing $I$ when collecting $X(s)$.

Thanks

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Thx Tong for fixing my question. –  OOzy Pal Apr 28 '13 at 4:26

$$sX(s)-x(0)=AX(s) \\ sX(s)-AX(s)=x(0) \\$$

In the second equation, notice that there is a factor of $X(s)$ on the right-hand side of both $sX(s)$ and $AX(s)$, so we can take it out like so:

$$(s-A)X(s)=x(0)$$

But this wouldn't make any sense, because $A$ is a matrix and $s$ is a scalar. To make it so that the equation makes sense, we make $s$ into a matrix by multiplying by the identity.

$$(sI-A)X(s)=x(0)$$

This is just to ensure that it doesn't matter which way round we do things. We could do $sI-A$ first, and then multiply the result by $X(s)$ on the right, or we can multiply through by $X(s)$ to get

$$sIX(s)-AX(s)=sX(s)-AX(s)$$

which is what you started with.

Adding the identity $I$ doesn't change the value of the equation, it's just so that it "makes sense" if we were to do the subtraction before the matrix multiplication.

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$$\begin{array}{rcl} \dot x&=Ax \\ sX(s)-x(0)&=AX(s) \\ sX(s)-AX(s)&=x(0) \\ sIX(s)-AX(s)&=x(0) \\ (sI-A)X(s)&=x(0) \end{array}$$

$I$ was already there but it wasn't written, because writing it doesn't change the meaning of the expression. Consider this analogical example from scalar equations:

$$\begin{array}{rcl} ab-a&=c \\ a(b-\underset{?}{\underset{\downarrow}{1}})&=c \end{array}$$

How did that $1$ appear there? It wasn't there, was it? Yes it was there, but it wasn't written, it was hidden, because it is the identity element of multiplication, just like $I$ is in matrix case.

$$\begin{array}{rcl} ab-a&=c \\ ab-a1&=c \\ a(b-1)&=c \end{array}$$

The hidden identity element comes into appearance when it is needed. In your case, it has to appear, because when you take the common factor $X(s)$ out of the parenthesis, you left $s-A$ inside. The dimensions must match but they don't. It was originally a difference of matrices $(s(X(s)-AX(s))$, but now a difference between a scalar and a matrix $(s-A)$ which is an undefined operation.

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Thank you hkBattousai for answering my question. Both Answers made sense. Thx –  OOzy Pal Apr 28 '13 at 4:29