3 and 4 .Determinants and Matrices
easy

$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$ માટે ગુણધર્મ $1$ ચકાસો. 

Option A
Option B
Option C
Option D

Solution

Solution Expanding the determinant along first row, we have

$\Delta  = 2\left| {\begin{array}{*{20}{r}}
  0&4 \\ 
  5&{ – 7} 
\end{array}} \right| – ( – 3)\left| {\begin{array}{*{20}{r}}
  6&4 \\ 
  1&{ – 7} 
\end{array}} \right| + 5\left| {\begin{array}{*{20}{r}}
  6&0 \\ 
  1&5 
\end{array}} \right|$

$ = 2(0 – 20) + 3( – 42 – 4) + 5(30 – 0)$

$ =  – 40 – 138 + 150 =  – 28$

By interchanging rows and columns, we get

$\Delta_{1}=\left|\begin{array}{rrr}2 & 6 & 1 \\ -3 & 0 & 5 \\ 5 & 4 & -7\end{array}\right| \quad$ (Expanding along first column)

$=2\left|\begin{array}{rr}
0 & 5 \\
4 & -7
\end{array}\right|-(-3)\left|\begin{array}{rr}
6 & 1 \\
4 & -7
\end{array}\right|+5\left|\begin{array}{ll}
6 & 1 \\
0 & 5
\end{array}\right|$

${ = 2(0 – 20) + 3( – 42 – 4) + 5(30 – 0)}$

${ =  – 40 – 138 + 150 =  – 28}$

Clearly $\Delta=\Delta_{1}$

Hence, Property $1$ is verified.

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.