Test Index

CBSE Class 12 Math 2024 All Sets Solved Paper

© examsnet.com
Question : 3 of 20
Marks: +1, -0
If ∣−abca−bcab−c∣=kabc\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc, then the value of k is
Solution:  
∣−abca−bcab−c∣=kabc\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc
abc∣−1111−1111−1∣=kabcabc\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = kabc (by taking a, b, c out of the matrix from column C1,C2,C3C_1, C_2, C_3 respectively)
∣−1111−1111−1∣=k\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = k (dividing by abc on both side)
∣−100102120∣=k\begin{vmatrix} -1 & 0 & 0 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{vmatrix} = k (by using C2←C2+C1C_2 \leftarrow C_2 + C_1 and C3←C3+C1C_3 \leftarrow C_3 + C_1)
−1(0×0−2×2)=k-1(0\times 0 - 2 \times 2) = k
−1(−4)=k-1(-4) = k
∴k=4\therefore k = 4
© examsnet.com
Go to Question:
Prev Question
Next Question