Q46 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $A X=D$ represents the system of linear equations $3 x-4 y+7 z+6=0,5 x+2 y-4 z+9=0$ and $8 x-6 y-z+5=0$, then
- A$\operatorname{Rank}(A)=\operatorname{Rank}([A D])=1$
- B$\operatorname{Rank}(A)=\operatorname{Rank}([A D])=2$
- Correct answerC$\operatorname{Rank}(A)=\operatorname{Rank}([A D])=3$
- D$\operatorname{Rank}\neq \operatorname{Rank}([A D])$
Solution
Here $A=\left[\begin{array}{ccc}3 & -4 & 7 \\ 5 & 2 & -4 \\ 8 & -6 & -1\end{array}\right] ; \quad|A|=\left|\begin{array}{ccc}3 & -4 & 7 \\ 5 & 2 & -4 \\ 8 & -6 & -1\end{array}\right|$
$\begin{aligned}
& =3(-2-24)+4(-5+32)+7(-30-16) \\
& =-78+108-322=-292 \neq 0 \quad \therefore \operatorname{Rank}(A)=3 \\
& A D=\left[\begin{array}{ccccc}
3 & -4 & 7 & : & -6 \\
5 & 2 & -4 & : & -9 \\
8 & -6 & -1 & : & -5
\end{array}\right]
\end{aligned}$
This is $3 \times 4$ matrix. So maximum rank of $A D$ is 3 and we know that $\left|\begin{array}{ccc}3 & -4 & 7 \\ 5 & 2 & -4 \\ 8 & -6 & -1\end{array}\right| \neq 0$ $\therefore \operatorname{Rank}(A D)=3$.
This is $3 \times 4$ matrix. So maximum rank of $A D$ is 3 and we know that $\left|\begin{array}{ccc}3 & -4 & 7 \\ 5 & 2 & -4 \\ 8 & -6 & -1\end{array}\right| \neq 0$ $\therefore \operatorname{Rank}(A D)=3$.
Hence, the correct answer is option: 'C'.