There are probably more sophisticated answers available here, but let me just mention one small way in which $K$-theory relates to linear algebra, which is that the lower $K$-groups successfully generalize the notion of a determinant.
Let $V$ be an f.g. projective module over a ring $R$. If $V$ is of rank $n$, then its $n$th exterior power, $\Lambda^{n} V$, is a rank 1 projective $ R $ module, i.e., an element of the Picard group. Call this element $\operatorname{det}(V) $. For a short exact sequence $ 0 \rightarrow V' \rightarrow V \rightarrow V'' \rightarrow0 $ there is \emph{canonical} isomorphism $\operatorname{det}(V) \cong \operatorname{det}(V') \otimes \operatorname{det}(V'')$, so the map extends to $K_{0} (R) \rightarrow \operatorname{Pic}(R)$. The determinant picks out the non-trivial part of $ K_{0}(R)$, the part which doesn't come from free modules.
Back to linear algebra. Since $\Lambda^{n} $ is a functor, if $ f:V \rightarrow V $ is a homomorphism, then $\operatorname{det} (f) := \Lambda^{n} (f)$ is in $\operatorname{End} (\Lambda^{n}V) \cong R $. In this way, the assertion $\operatorname{det}(fg) = \operatorname{det} (f) \operatorname{det}(g)$, usually a pain to prove to undergraduates, follows trivially. The map $ GL (R) \rightarrow R^{\times} $, $A \mapsto \operatorname{det}(A) $ descends to a surjection $ K_{1} (R) \rightarrow R^{\times} $, which is an isomorphism if $ R $ is a commutative local ring. The failure of this map to be an isomorphism is in some sense a measure of the failure of some parts of linear algebra over $ R $.