Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$
(N/A) The determinant is given by $\Delta = \left| \begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array} \right|$.
Expanding along the first row:
$\Delta = 2(0 - 20) - (-3)(-42 - 4) + 5(30 - 0) = 2(-20) + 3(-46) + 5(30) = -40 - 138 + 150 = -28$.
Interchanging rows $R_{2}$ and $R_{3}$ i.e.,$R_{2} \leftrightarrow R_{3}$,we get $\Delta_{1} = \left| \begin{array}{ccc} 2 & -3 & 5 \\ 1 & 5 & -7 \\ 6 & 0 & 4 \end{array} \right|$.
Expanding $\Delta_{1}$ along the first row:
$\Delta_{1} = 2(20 - 0) - (-3)(4 + 42) + 5(0 - 30) = 2(20) + 3(46) + 5(-30) = 40 + 138 - 150 = 28$.
Clearly,$\Delta_{1} = -\Delta$,which is $28 = -(-28)$.
Hence,Property $2$ is verified.
Expanding along the first row:
$\Delta = 2(0 - 20) - (-3)(-42 - 4) + 5(30 - 0) = 2(-20) + 3(-46) + 5(30) = -40 - 138 + 150 = -28$.
Interchanging rows $R_{2}$ and $R_{3}$ i.e.,$R_{2} \leftrightarrow R_{3}$,we get $\Delta_{1} = \left| \begin{array}{ccc} 2 & -3 & 5 \\ 1 & 5 & -7 \\ 6 & 0 & 4 \end{array} \right|$.
Expanding $\Delta_{1}$ along the first row:
$\Delta_{1} = 2(20 - 0) - (-3)(4 + 42) + 5(0 - 30) = 2(20) + 3(46) + 5(-30) = 40 + 138 - 150 = 28$.
Clearly,$\Delta_{1} = -\Delta$,which is $28 = -(-28)$.
Hence,Property $2$ is verified.
Explore More
Similar Questions
If $a=1+2+4+\cdots$ up to $n$ terms,$b=1+3+9+\cdots$ up to $n$ terms and $c=1+5+25+\cdots$ up to $n$ terms,then $\Delta=\left|\begin{array}{ccc}a & 2b & 4c \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n\end{array}\right|=$
$A$ value of $\theta \in (0, \pi /3)$,for which $\left| \begin{array}{ccc} 1 + \cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{array} \right| = 0$,is
DifficultJEE MAIN 2019
View SolutionIf $x, y$ and $z$ are greater than $1$,then the value of $\left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$ is
MediumWBJEE 2016
View Solution$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $
Easy
View SolutionIf $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$,then $\sin 4\theta$ is equal to:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo