If $\left| {\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}} \right| = 0$,then $a, b, c$ are in:

  • A
    $A.P.$
  • B
    $G.P.$
  • C
    $H.P.$
  • D
    None of these

Explore More

Similar Questions

If ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$ for all $a, b, c \in R$,then find the value of the determinant $\left| {\begin{array}{*{20}{c}} {{(a + b + c)}^2} & {{a^2} + {b^2}} & 1 \\ 1 & {{(b + c + 2)}^2} & {{b^2} + {c^2}} \\ {{c^2} + {a^2}} & 1 & {{(c + a + 2)}^2} \end{array}} \right|$.

Difficult
View Solution

The sum of the distinct values of $x$ for which the matrix $A=\begin{bmatrix} 1 & 1 & x \\ 1 & x & 1 \\ x & 1 & 1 \end{bmatrix}$ has no inverse,is

If $A, B, C$ are the angles of a triangle,then $\left| \begin{array}{ccc} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{array} \right| = $

Difficult
View Solution

Evaluate the determinant: $\left| \begin{array}{ccc} 1 & a & b \\ -a & 1 & c \\ -b & -c & 1 \end{array} \right|$

If $k > 1$ and the determinant of the matrix $A^2$, where $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$, is $k^2$, then $|\alpha|$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo