Let $A \in \mathrm{Mat}_{n n}$. Then $\operatorname{det} A=0$ if and only if $\operatorname{ker} L_A \neq\{\underline{0}\}$. Proof. Let the RREF of $A$ be $E A$, where $E$ is a product of invertible elementary matrices. Either $E A=I_n$ (and $\operatorname{det} E A=1$ ) or $E A$ has at least one zero row (and $\operatorname{det} E A=0$ ). In the first case $\operatorname{ker} L_{E A}=\{\underline{0}\}$, while in the second case $\operatorname{dim} \operatorname{ker} L_{E A} \geq 1$. Therefore $\operatorname{det} E A=0$ if and only if $\operatorname{dim} \operatorname{ker} L_{E A} \geq 1$. The proof is completed by recalling that $\operatorname{ker} L_{E A}=\operatorname{ker} L_A$, either by [[Elementary Row Operations Preserve Kernel of Left Multiplication Linear Map of Real Matrix]] or by recalling how row operations correspond to solving linear equations.