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3 and 4 .Determinants and Matrices
normal
Let $ABC = I$ then $tr(ABC + BCA + CAB)$ is (where order of matrices $A, B, C$ is $3$ and $tr(A)$ is sum of diagonal elements in $A$)
A
$3$
B
$9$
C
$12$
D
$15$
Solution
$\mathrm{ABC}=\mathrm{I}$
$\mathrm{BC}=\mathrm{A}^{-1}$
$\mathrm{BCA}=\mathrm{I}$
$\mathrm{BC}=\mathrm{A}^{-1}$
$\mathrm{C}=\mathrm{B}^{-1} \mathrm{A}^{-1}$
$\mathrm{CA}=\mathrm{B}^{-1} \Rightarrow \mathrm{CAB}=\mathrm{I}$
$\operatorname{tr}(3 \mathrm{I})=9$
Standard 12
Mathematics