Certainly! Let's clarify the statement. You are asked to prove that if a block matrix \( M \) can be written as \[ M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} \] where **A** and **C** are square matrices (the blocks on the diagonal), then \[ \det(M) = \det(A) \cdot \det(C) \] Let's solve this step by step: --- ### **Step 1: Understand the Structure of the Block Matrix** Given: \[ M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} \] where \(A\) is \(k \times k\), \(C\) is \(l \times l\), and \(B\) is \(k \times l\). **Explanation:** This is an upper block triangular matrix because the lower-left block is zero. We are interested in the determinant of such a matrix. --- ### **Step 2: Recall Property of Determinant for Block Upper Triangular Matrices** For any block upper triangular matrix of the form: \[ \begin{pmatrix} P & Q \\ 0 & R \end{pmatrix} \] where \(P\) and \(R\) are square matrices (of any size), **the determinant is given by:** \[ \det \begin{pmatrix} P & Q \\ 0 & R \end{pmatrix} = \det(P)\cdot\det(R) \] **Explanation:** This property holds because, when expanding the determinant, the only nonzero contributions come from products that pick entries from the diagonal blocks, just as in the case of ordinary triangular matrices. The off-diagonal block \(Q\) does not affect the determinant in this structure. --- ### **Step 3: Apply the Property to Your Matrix** Apply the property directly: \[ \det(M) = \det \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} = \det(A) \cdot \det(C) \] **Explanation:** No matter what \(B\) is, as long as the lower-left block is zero and the diagonal blocks are square, this formula holds. The key is that the lower left block is zero, making the matrix block upper triangular. --- ## **Final Answer** \[ \boxed{ \det \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} = \det(A)\cdot\det(C) } \] **This holds even if the off-diagonal block \(B\) is not zero, as long as the lower-left block is zero and \(A, C\) are square matrices.**