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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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:

Similar Questions

Solve the equation $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$,where $a \neq 0$.

If $5$ is one root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{array} \right| = 0$,then the other two roots of the equation are:

One of the roots of the given equation $\left| \begin{array}{ccc} x+a & b & c \\ b & x+c & a \\ c & a & x+b \end{array} \right| = 0$ is

The roots of the equation $\left| \begin{matrix} x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x \end{matrix} \right| = 0$ are equal to

If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module