Test Index

CBSE Class 12 Math 2022 Term I Solved Paper

© examsnet.com
Question : 13 of 50
Marks: +1, -0
If CijC_{ij} denotes the cofactor of element PijP_{ij} of the matrix P=[1−1202−3324]P=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & 2 & 4 \end{bmatrix}, then the value of C31⋅C23C_{31} \cdot C_{23} is
Solution:     👈: Video Solution
Explanation:
Here,
C31=(−1)3+1∣−122−3∣=3−4=−1C_{31}=(-1)^{3+1}\begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}=3-4=-1
C23=(−1)2+3∣1−132∣=−(2+3)=−5C_{23}=(-1)^{2+3}\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}=-(2+3)=-5
and
C23=(−1)2+3∣1−132∣=−(2+3)=−5C_{23}=(-1)^{2+3}\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}=-(2+3)=-5
Thus,     C31⋅C23=(−1)(−5)=5\; \; C_{31} \cdot C_{23}=(-1)(-5)=5
© examsnet.com
Go to Question:
Prev Question
Next Question