Question

If $a, b, c$ are real numbers such that $a-b=1, b-c=3$, 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

• A. 1
• B. 2
• C. 3
• D. infinitely many

Answer 1

Answer: (D)

We have,
$A=\begin{bmatrix}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{bmatrix}$
$|A|=\begin{bmatrix}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{bmatrix}$
$|A|=\begin{bmatrix}1 & 0 & 0 \\ a & b-a & c-a \\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2}\end{bmatrix}=(b-a)(c-a)(c-b)$
$|A|=(a-b)(b-c)(c-a)$
$|A|=-12,(a-b)=1,(b-c)=3$
$\because\,(a-b)(b-c)(c-a)=-12$
$ 1 \times 3 \times(c-a)=-12$
$c-a=-4 $
$\Rightarrow \, a-c=4$
$\because \,a-b=1, b-c=3, a-c=4$
$\therefore $ Infinite solution.