(C) Given the determinant: $\Delta = \begin{vmatrix} 2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1 \end{vmatrix}$.Expanding along the first row:$\Delta = 2

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

(C) Given the determinant: $\Delta = \begin{vmatrix} 2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1 \end{vmatrix}$.Expanding along the first row:$\Delta = 2[(k-1)(k+1) - 1] - 2k[1(k+1) - 2] + 1[1 - 2(k-1)]$$\Delta = 2[k^2 - 1 - 1] - 2k[k + 1 - 2] + 1[1 - 2k + 2]$$\Delta = 2[k^2 - 2] - 2k[k - 1] + [3 - 2k]$$\Delta = 2k^2 - 4 - 2k^2 + 2k + 3 - 2k$$\Delta = -1$.Comparing $\Delta = -1$ with $Ak^2 + Bk + C$, we get $A = 0, B = 0, C = -1$.Therefore, $A + B + C = 0 + 0 - 1 = -1$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Easy

If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

TS EAMCET 2021 All TS EAMCET

A
$0$
B
$1$
C
$-1$
D
$2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3-4. Matrices and Determinants

Subtopic

Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Previous Year

TS EAMCET 2021 Paper All TS EAMCET years →

$\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| = $

Medium

View Solution

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:

MediumAP EAMCET 2012

View Solution

If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.

DifficultAP EAMCET 2020

View Solution

$\left| {\begin{array}{ccc} 1 + i & 1 - i & i \\ 1 - i & i & 1 + i \\ i & 1 + i & 1 - i \end{array}} \right| = $

Medium

View Solution

If $a, b, c$ are distinct positive real numbers,then the value of the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is

EasyAP EAMCET 2016

View Solution

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

If $\left|\begin{array}{ccc}2 & 2k & 1 \\ 1 & k-1 & 1 \\ 2 & 1 & k+1\end{array}\right|=Ak^2+Bk+C$, then $A+B+C=$

Similar Questions

$\left| {\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| = $

If $f(x) = \left|\begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & (x-1)x(x+1) \end{array}\right|$, then $f(2012)$ is equal to:

If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.

$\left| {\begin{array}{ccc} 1 + i & 1 - i & i \\ 1 - i & i & 1 + i \\ i & 1 + i & 1 - i \end{array}} \right| = $

If $a, b, c$ are distinct positive real numbers,then the value of the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module