= Solution
Use the usual unital convention for an <algebra over a field>. For $a,b,x\in A$ and $c\in K$,
$$
\phi_{a+cb}(x)=\phi_a(x)+c\phi_b(x),\qquad
\phi_{ab}(x)=a(bx)=(\phi_a\circ\phi_b)(x),\qquad \phi_1(x)=x.
$$
Passing to matrices in one fixed <basis> preserves these identities: $M_{a+cb}=M_a+cM_b$, $M_{ab}=M_aM_b$ and $M_1=I$. Thus $\psi$ is an <algebra homomorphism>. If $M_a=0$, evaluating $\phi_a$ at $1_A$ gives $a=0$. Its <kernel> is zero, so \b[the left regular representation is faithful].
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