Let's solve each part step by step with explanations: --- ### (a) Prove that \(\det(EA) = \det(E)\det(A)\) #### **Step 1:** Recall that an **elementary matrix** \( E \) represents a single elementary row operation. Multiplying any matrix \( A \) by \( E \) on the left, \( EA \), performs that elementary row operation on \( A \). #### **Step 2:** By properties of determinants (from earlier questions or standard results), multiplying a matrix \( A \) by an elementary matrix \( E \) multiplies \( \det(A) \) by \( \det(E) \): \[ \det(EA) = \det(E)\det(A) \] **Explanation:** This holds because the determinant is a multilinear, alternating function, and each elementary row operation (scaling, swapping, or adding rows) has a predictable effect on the determinant, exactly as encoded by \( \det(E) \). --- ### (b) Prove that if \( A \) is not invertible, then \( AB \) is not invertible. #### **Step 1:** If \( A \) is **not invertible**, then \( \det(A) = 0 \). #### **Step 2:** The product \( AB \) is invertible **if and only if** \( \det(AB) \neq 0 \). But from the multiplicative property of determinants: \[ \det(AB) = \det(A)\det(B) \] If \( \det(A) = 0 \), then \( \det(AB) = 0 \), so \( AB \) is **not invertible**. **Explanation:** Since the determinant of a product is zero when one factor's determinant is zero, \( AB \) cannot be invertible if \( A \) isn't. --- ### (c) If \( A \) is invertible, show it can be written as a product of elementary matrices, and relate their determinants. #### **Step 1:** **Fact:** Any invertible matrix \( A \) can be reduced to the identity matrix by a sequence of elementary row operations. Thus, \( A \) can be written as a product of elementary matrices: \[ A = E_1^{-1} E_2^{-1} \cdots E_k^{-1} \] Or, equivalently, \[ A = F_1 F_2 \cdots F_k \] where each \( F_i \) is an elementary matrix. #### **Step 2:** The determinant of a product is the product of determinants: \[ \det(A) = \prod_{i=1}^k \det(F_i) \] **Explanation:** Since each elementary matrix represents a row operation, and the determinant is multiplicative over matrix products, the determinant of \( A \) is the product of the determinants of these elementary matrices. --- ### (d) Use (a), (b), and (c) to prove \(\det(AB) = \det(A)\det(B)\). #### **Step 1:** If \( A \) is **not invertible** (\(\det(A) = 0\)), then by (b), \( AB \) is not invertible (\(\det(AB) = 0\)), and so \[ \det(AB) = 0 = 0 \cdot \det(B) = \det(A)\det(B) \] #### **Step 2:** If \( A \) **is invertible**, by (c), \( A \) can be written as a product of elementary matrices, \( A = E_1 E_2 \cdots E_k \). Thus, \[ AB = (E_1 E_2 \cdots E_k)B \] By repeated application of (a), \[ \det(AB) = \det(E_1(E_2(\cdots(E_k B)\cdots))) = \prod_{i=1}^k \det(E_i)\det(B) = \det(A)\det(B) \] **Final Answer:** \[ \boxed{\det(AB) = \det(A)\det(B)} \] This holds whether \( A \) is invertible or not.

Let's systematically work through each part step by step, providing clear explanations. --- ### (a) Prove that \(\det(EA) = \det(E) \det(A)\) **Step 1:** Recall that an elementary matrix \(E\) represents a single elementary row operation. When we multiply \(A\) on the left by \(E\), we perform that row operation on \(A\). **Step 2:** From properties of determinants, multiplying a matrix \(A\) by an elementary matrix \(E\) on the left scales the determinant by \(\det(E)\): \[ \det(EA) = \det(E) \times \det(A) \] **Explanation:** Elementary row operations correspond to elementary matrices, and their effect on determinants is well known: - Swapping rows multiplies the determinant by \(-1\) (\(\det(E) = -1\)). - Scaling a row multiplies the determinant by the scaling factor (\(\det(E) = \text{scaling factor}\)). - Adding multiples of one row to another does not change the determinant (\(\det(E) = 1\)). Thus, the property holds generally for elementary matrices. --- ### (b) Prove that if \(A\) is not invertible, then \(AB\) is not invertible. **Step 1:** If \(A\) is **not invertible**, then \(\det(A) = 0\). **Step 2:** The determinant of a product satisfies: \[ \det(AB) = \det(A) \times \det(B) \] Since \(\det(A) = 0\), it follows: \[ \det(AB) = 0 \times \det(B) = 0 \] which implies \(AB\) is **not invertible** because its determinant is zero. **Explanation:** The invertibility of a matrix depends on its determinant being non-zero. If \(A\) isn't invertible, multiplying it by any matrix \(B\) cannot produce an invertible matrix. --- ### (c) If \(A\) is invertible, show that it can be written as a product of elementary matrices, and relate their determinants. **Step 1:** Any invertible matrix \(A\) can be obtained from the identity matrix \(I\) through a finite sequence of elementary row operations. Therefore, \[ A = E_1 E_2 \cdots E_k \] or, equivalently, \(A\) can be expressed as a product of elementary matrices \(E_i\). **Step 2:** Since the determinant function is multiplicative over matrix products, \[ \det(A) = \det(E_1) \times \det(E_2) \times \cdots \times \det(E_k) \] **Explanation:** Each elementary matrix corresponds to a specific row operation with a known effect on the determinant. The product of their determinants gives the overall effect, matching the determinant of \(A\). --- ### (d) Use (a), (b), and (c) to prove that \(\det(AB) = \det(A)\det(B)\). **Step 1:** - If \(A\) is **not invertible**, then \(\det(A) = 0\), and by (b), \(AB\) is **not invertible** (\(\det(AB) = 0\)). Therefore, \[ \det(AB) = 0 = \det(A) \times \det(B) \] - If \(A\) **is invertible**, from (c), \(A\) can be expressed as a product of elementary matrices: \[ A = E_1 E_2 \cdots E_k \] Thus, \[ AB = (E_1 E_2 \cdots E_k) B \] **Step 2:** Applying (a), the determinant of the product is: \[ \det(AB) = \det(E_1) \det(E_2) \cdots \det(E_k) \times \det(B) = \det(A) \times \det(B) \] since \(\det(A) = \prod_{i=1}^k \det(E_i)\). **Final conclusion:** \[ \boxed{ \det(AB) = \det(A) \times \det(B) } \] This proof holds regardless of whether \(A\) is invertible or not, completing the argument. --- **Final answer:** \[ \boxed{ \text{For all square matrices } A, B,\quad \det(AB) = \det(A) \det(B) } \]

Let's go through each part carefully with explanations, then provide a final summary at the end. --- ### (a) Prove that \(\det(EA) = \det(E) \det(A)\) **Step 1:** Recall that an elementary matrix \(E\) corresponds to a single elementary row operation. Multiplying \(A\) on the left by \(E\) results in the matrix \(EA\), which is the matrix \(A\) after applying that row operation. **Step 2:** From properties of determinants (either from prior knowledge or earlier exercises), we know that: \[ \det(EA) = \det(E) \times \det(A) \] **Explanation:** Elementary row operations affect the determinant in predictable ways: - Row swapping: \(\det(E) = -1\), and \(\det(EA) = -\det(A)\). - Row scaling: \(\det(E) = \text{scaling factor}\), and \(\det(EA) = \text{scaling factor} \times \det(A)\). - Row addition: \(\det(E) = 1\), and \(\det(EA) = \det(A)\). Thus, the general rule holds that multiplying by an elementary matrix scales the determinant by \(\det(E)\). --- ### (b) Prove that if \(A\) is not invertible, then \(AB\) is not invertible **Step 1:** If \(A\) is **not invertible**, then \(\det(A) = 0\). **Step 2:** Using the determinant property for products: \[ \det(AB) = \det(A) \times \det(B) \] Since \(\det(A) = 0\), it follows: \[ \det(AB) = 0 \times \det(B) = 0 \] which means \(AB\) is **not invertible** (since a matrix is invertible iff its determinant is non-zero). **Explanation:** The zero determinant of \(A\) guarantees the product \(AB\) cannot be invertible regardless of \(B\). --- ### (c) If \(A\) is invertible, show that it can be written as a product of elementary matrices, and relate their determinants. **Step 1:** Any invertible matrix \(A\) can be obtained from the identity matrix \(I\) via a finite sequence of elementary row operations. Each of these operations corresponds to an elementary matrix \(E_i\), and thus: \[ A = E_1 E_2 \cdots E_k \] **Step 2:** The determinant of \(A\) is the product of the determinants of these elementary matrices: \[ \det(A) = \prod_{i=1}^k \det(E_i) \] **Explanation:** Since each elementary matrix has a known determinant (\(\pm 1\) or a scalar), expressing \(A\) as a product of elementary matrices makes the calculation of \(\det(A)\) straightforward via multiplication of those determinants. --- ### (d) Use (a), (b), and (c) to prove that \(\det(AB) = \det(A)\det(B)\) **Step 1:** - If \(A\) is **not invertible** (\(\det(A) = 0\)), then from (b), \(AB\) is not invertible (\(\det(AB)=0\)). So, \[ \det(AB) = 0 = \det(A) \times \det(B) \] - If \(A\) **is invertible**, from (c), \(A\) can be written as a product of elementary matrices \(E_i\): \[ A = E_1 E_2 \cdots E_k \] Thus, \[ AB = (E_1 E_2 \cdots E_k) B \] **Step 2:** Applying (a), the determinant of a product is: \[ \det(AB) = \det(E_1) \times \det(E_2) \times \cdots \times \det(E_k) \times \det(B) = \det(A) \times \det(B) \] since \(\det(A) = \prod_{i=1}^k \det(E_i)\). **Final conclusion:** \[ \boxed{ \det(AB) = \det(A) \det(B) } \] **This completes the proof for all cases, whether \(A\) is invertible or not.** --- ### **Final summarized answer:** \[ \boxed{ \text{For all square matrices } A, B,\quad \det(AB) = \det(A) \det(B) } \] This fundamental property of determinants holds generally and is consistent with the properties of elementary matrices and their effects on determinants.