give answer in 2 step with explanation at the end of each step and final answer at the end:Question 3: 6 Marks
Let D and E be two square matrices of the same size such that det(E) 0 and det(E) # 0. P

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Question 3: 6 Marks
Let D and E be two square matrices of the same size such that det(E) 0 and det(E) # 0. Prove that the
product DE has an inverse and

ED" (DE)

VIPSolutions · Accepted Answer

**Step 1: Prove that $ DE $ has an inverse**

Since $ D $ and $ E $ are square matrices of the same size and both have nonzero determinants ($\det(D) \neq 0$, $\det(E) \neq 0$), both are invertible matrices.  
Recall: The product of two invertible matrices is also invertible, and the determinant of their product is the product of their determinants:
$$
\det(DE) = \det(D)\det(E) \neq 0
$$
Thus, $ DE $ is invertible.

**Explanation:**  
A square matrix is invertible if and only if its determinant is nonzero. Since both $ D $ and $ E $ are invertible, so is their product $ DE $.

---

**Step 2: Show that $ E^{-1}D^{-1} = (DE)^{-1} $**

Recall the property of inverses for the product of two invertible matrices:
$$
(AB)^{-1} = B^{-1}A^{-1}
$$
Applying this to $ DE $:
$$
(DE)^{-1} = E^{-1}D^{-1}
$$
So,
$$
E^{-1}D^{-1} = (DE)^{-1}
$$

**Explanation:**  
The inverse of a product of invertible matrices is the product of their inverses in the reverse order.

---

**Final Answer:**

- **$ DE $ is invertible.**
- **$ E^{-1}D^{-1} = (DE)^{-1} $**.

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give answer in 2 step with explanation at the end of each step and final answer at the end:Question 3: 6 Marks Let D and E be two square matrices of the same size such that det(E) 0 and det(E) # 0. Prove that the product DE has an inverse and ED" (DE)

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