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An upper triangular matrix is a matrix where all the elements below its main diagonal are zeros, i.e., the matrix A of size n x n is an upper triangular matrix if a_{ij} = 0 for all i > j. An example of a 3 x 3 upper triangular matrix: \(A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ 0 & a_{22} & a_{23}\\ 0 & 0 & a_{33}\end{bmatrix}\)

Let's compute the determinant of the above 3 x 3 upper triangular matrix A using cofactor expansion. The determinant of a 3 x 3 matrix A: \(|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}\) where C_{ij} is the cofactor of the entry a_{ij}. Now, let's calculate the cofactors: \(C_{11} = |A_{11}| = \begin{vmatrix} a_{22} & a_{23}\\ 0 & a_{33}\end{vmatrix} = a_{22}a_{33}\) \(C_{12} = (-1)^{1+2}|A_{21}| = -\begin{vmatrix} 0 & a_{23}\\ 0 & a_{33}\end{vmatrix} = 0\) \(C_{13} = (-1)^{1+3}|A_{31}| = \begin{vmatrix} 0 & a_{22}\\ 0 & 0\end{vmatrix} = 0\) Plugging the cofactors back into the determinant formula, we get: \(|A| = a_{11}(a_{22}a_{33}) + 0 + 0 = a_{11}a_{22}a_{33}\)

Now that we have seen the proof for a 3 x 3 upper triangular matrix, we can generalize it for an n x n upper triangular matrix. We will prove it by induction: The base case of a 1 x 1 matrix is already trivially true, since the determinant of a matrix with a single element is simply the value of the single element. Now let's assume that the determinant of an upper triangular matrix of size (n-1) x (n-1) is the product of its diagonal elements: \(|A_{n-1}| = a_{11}a_{22}...a_{(n-1)(n-1)}\) Now consider an upper triangular matrix A of size n x n. Using cofactor expansion, we have: \(|A| = a_{11}C_{11} + a_{12}C_{12} + ... + a_{1n}C_{1n}\) As we proceed through the cofactors, we find that the cofactor of a_{1k} is zero for k>1 because the column contains a zero in the position (2, k). Therefore: \(|A| = a_{11}C_{11}\) Reviewing the definition of the determinant as the product of the diagonal elements of the submatrices we can write: \(|A| = a_{11}|A_{11}|\) Now, because we assumed that the determinant of an upper triangular matrix of size (n-1) x (n-1) is the product of its diagonal elements, we have: \(|A_{11}| = a_{22}a_{33}...a_{nn}\) Hence, the determinant of the upper triangular matrix A is the product of its diagonal elements: \(|A| = a_{11}a_{22}a_{33}...a_{nn}\) This concludes the proof that the determinant of an upper triangular matrix is the product of its diagonal entries.