GATE | GATE MOCK 2017 | Question 14

mock

(A) 1
(B) 2
(C) 3
(D) 0


Answer: (D)

Explanation: Applying R_{2}\rightarrow R_{2} + 2R_{1}; R_{3}\rightarrow R_{3} + 3R_{1}; R_{4}\rightarrow R_{4} + 5R_{1};
Q \sim \begin{bmatrix}  -1& 2 & 3 &-2 \\    0& -1 & 7 & -2\\    0& -2 & 14 & -4\\    0& -2 & 14 & -4 \end{bmatrix}
Applying R_{3}\rightarrow R_{3} - 2R_{2}; R_{4}\rightarrow R_{4} + 2R_{2}
Q \sim \begin{bmatrix}  -1& 2 & 3 &-2 \\    0& -1 & 7 & -2\\    0&  0 & 0 &  0\\    0&  0 & 0 & 0 \end{bmatrix}
\therefore It is echelon form.
\therefore Rank of Q = 2 < n-1 We know that if \rho (A_{n \times n}) < n-1 then \rho (Adj \ A) = 0.
So, Rank of Adjoint matrix of Q is 0.

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