Determinant of the Identity Matrix proof

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I have trouble proving that for all $n$, $\det(I_{n})=1$

$I_{n}$ is Identity Matrix $nxn$

I tried to use Inductive reasoning but without any progress

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1 Answer

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For $\;n=1\;$ we clearly have $\;\det (1)=1\;$ , and even directly for $\;n=2\;$ :

$$\det\begin{pmatrix}1&0\\0&1\end{pmatrix}=1\cdot\det(1)=1$$

Now, take $\;I_n\;$ and develop with respect the first row (or the first column, it is exactly the same), then you get:

$$\det I_n=1\cdot \det I_{n-1}\stackrel{\text{inductive hypotesis}}=1\cdot1=1$$

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