Question

If $A$ and $B$ are square matrices and of the same order such that $A^{2024} = 0$ and $B = AB - A$ then det $(B)$ is

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

Answer: (B)

0

Answer 2

Given that $A$ and $B$ are square matrices of the same order.

We are given two conditions:

1. $A^{2024} = 0$
2. $B = AB - A$

Step 1: Analyze the first condition $A^{2024} = 0$. Taking the determinant of both sides of the equation: $\det(A^{2024}) = \det(0)$

The determinant of a zero matrix (a matrix with all elements being zero) is 0. Also, a property of determinants states that $\det(X^k) = (\det(X))^k$. Applying these properties: $(\det(A))^{2024} = 0$

For $(\det(A))^{2024}$ to be 0, $\det(A)$ must be 0. So, $\det(A) = 0$.

Step 2: Analyze the second condition $B = AB - A$. We need to find $\det(B)$. Let's rearrange the given equation: $B = AB - A$ We can factor out $A$ from the terms on the right-hand side. To do this, we can think of $A$ as being multiplied by the identity matrix $I$ when it stands alone, i.e., $A = AI$. $B = A(B - I)$

Step 3: Take the determinant of both sides of the rearranged equation. $\det(B) = \det(A(B - I))$

Using the property of determinants that $\det(XY) = \det(X) \det(Y)$ for square matrices $X$ and $Y$ of the same order: $\det(B) = \det(A) \det(B - I)$

Step 4: Substitute the value of $\det(A)$ found in Step 1. We found that $\det(A) = 0$. Substitute this into the equation: $\det(B) = 0 \cdot \det(B - I)$ $\det(B) = 0$

Thus, the determinant of matrix $B$ is 0.