Exercise 4.2.23

Use induction on $n$, the size of the matrix. For $n=1$, every $1\times 1$ matrix is upper triangular and we have the fact $\det <!--\; FUNCTION\; APPLICATION\; -->\left(\begin{array}{c}\hfill a\hfill \end{array}\right)=a$. Assuming the statement of this exercise holds for $n=k$, consider any $(n+1)\times (n+1)$ upper triangular matrix $A$. We can expand $A$ along the first row with the formula

And the matrix ${Ã}_{1j}$, $j\ne 1$, contains one zero column and hence has rank less than $n+1$. By the Corollary after Theorem 4.6 those matrix has determinant $0$. However, we have the matrix ${Ã}_{11}$ is upper triangular and by induction hypothesis we have

So we know the original formula would be