If a, b, c are real numbers such that a-b=1 a | vedclass

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(A) The determinant of the matrix $A$ is given by the Vandermonde determinant formula:
$|A| = \begin{vmatrix} 1 & 1 & 1 \ a & b & c

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$0$

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(A) The determinant of the matrix $A$ is given by the Vandermonde determinant formula:
$|A| = \begin{vmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{vmatrix} = (b-a)(c-a)(c-b)$.
We can rewrite this as $|A| = -(a-b)(c-a)(b-c)$.
Given $a-b=1$ and $b-c=3$, we have $c-b = -3$.
Also, $(a-c) = (a-b) + (b-c) = 1 + 3 = 4$, so $(c-a) = -4$.
Substituting these values into the determinant expression:
$|A| = -(1) 	imes (-4) 	imes (3) = 12$.
However, the problem states $|A| = -12$.
Since the value of the determinant is fixed at $12$ based on the given constraints $a-b=1$ and $b-c=3$, it can never be $-12$.
Therefore, there are $0$ such matrices.

If $a, b, c$ are real numbers such that $a-b=1$ and $b-c=3$, then the number of matrices of the form $A=\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ such that $|A|=-12$ is:

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