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3 and 4 .Determinants and Matrices
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If $A$, $B$ and $C$ are square matrices of order $3$ such that $A = \left[ {\begin{array}{*{20}{c}} x&0&1 \\ 0&y&0 \\ 0&0&z \end{array}} \right]$ and $\left| B \right| = 36$, $\left| C \right| = 4$, $\left( {x,y,z \in N} \right)$ and $\left| {ABC} \right| = 1152$ then the minimum value of $x + y + z$ is
A
$6$
B
$8$
C
$10$
D
$12$
Solution
$|A|=x y z$
$|A B C|=x y z \times 36 \times 4=1152$
$x y z=8$
$\frac{x+y+z}{3} \geq \sqrt[3]{8}$
$x+y+z \geq 6$
Standard 12
Mathematics
