Use the Leibniz formula to show that the matrix \(\mathbf{A}\in M_{5,5}(\mathbb{K})\) given by \[\mathbf{A}=\begin{pmatrix} * & * & * & * & * \\ * & * & * & * & * \\ * & * & 0 & 0 & 0 \\ * & * & 0 & 0 & 0 \\ * & * & 0 & 0 & 0 \end{pmatrix}\] has determinant zero. Here \(*\) stands for any scalar from \(\mathbb{K}.\)

Let \(n \in \mathbb{N}.\) Recall that for an upper triangular matrix \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n} \in M_{n,n}(\mathbb{K}),\) we have \[\det(\mathbf{A})=\prod_{i=1}^n A_{ii}.\] Prove this formula using the Leibniz formula.

Let \(n \in \mathbb{N},\) \(\mathbf{A}\in M_{n,n}(\mathbb{K})\) and \(\hat{\mathbf{A}}=(\hat{A}_{ij})_{1\leqslant i,j\leqslant n}\) be a REF matrix obtained from \(\mathbf{A}\) without multiplying with elementary matrices of the type \(\mathbf D_k(s).\) Show that \[\det(\mathbf{A})=(-1)^j \prod_{i=1}^n \hat{A}_{ii},\] where \(j\) is the number of exchange of rows during the reduction from \(\mathbf{A}\) to \(\hat{\mathbf{A}}.\)

Let \(\mathbf{A}\) be a \(2\)-by-\(2\) matrix with real entries and let \(\vec{a}_1,\vec{a}_2\) denote its columns. Consider the parallelogram \(P\) with corners \(0_{\mathbb{R}^2},\vec{a}_1,\vec{a}_2,\vec{a}_1+\vec{a}_2.\) Show that the surface area of \(P\) is given by the absolute value of the determinant of \(\mathbf{A}.\)