Q94 / 160 · AP EAMCET 2012 · Mathematics

Let $A=\left[\begin{array}{rrr}-1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right], B=\left[\begin{array}{cr}1 & -2 \\ -1 & 2\end{array}\right] \quad$ and $C=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$, if $a, b$ and $c$ respectively, denote the ranks of $A, B$ and $C$, then the correct order of these number is

  • A
    $a < b < c$
  • B
    $c < b < a$
  • Correct answer
    C
    $b < a < c$
  • D
    $a < c < b$

Solution

Given, $A=\left[\begin{array}{rrr}-1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]$ $\begin{aligned} \therefore|A| & =-1(24-25)+2(18-20)-3(15-16) \\ & =1-4+3=0\end{aligned}$ Now, $\left|\begin{array}{ll}4 & 5 \\ 5 & 6\end{array}\right|=24-25=-1 \neq 0$ $\therefore$ Rank of $A=a=2$ $B=\left[\begin{array}{rr}1 & -2 \\ -1 & 2\end{array}\right]$ $\therefore \quad|B|=2-2=0$ $\therefore$ Rank of $B=b=1$ and $\quad C=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$ $|C|=2(4-0)=8 \neq 0$ $\therefore$ Rank of $C=3, c=3$ $b < a < c \quad(\because 1 < 2 < 3)$

Hence, the correct answer is option: 'C'.

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