linearalgebra

Question

if $A = \begin{bmatrix} 2 &3 &4 \\ 3 & -1 &2 \\ -1& 4 & 5 \end{bmatrix}$ then rank of the matrix $(A-A^T)$ is _____

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

Comments

(B) 2

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i got 1 he gave 0

Answer 1

$A=\begin{bmatrix} 2 &3 &4 \\ 3 &-1 &2 \\ -1 & 4 & 5 \end{bmatrix}$               $A^{T}=\begin{bmatrix} 2 &3 &-1 \\ 3 &-1 &4 \\ 4 & 2 & 5 \end{bmatrix}$

$A-A^{T}=\begin{bmatrix} 2 &3 &4 \\ 3 &-1 &2 \\ -1 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 2 &3 &-1 \\ 3 &-1 &4 \\ 4 & 2 & 5 \end{bmatrix}$

$A-A^{T}=\begin{bmatrix} 0 &0 &5 \\ 0 & 0 & -2\\ -5 & 2 & 0 \end{bmatrix}$

$R2 \rightarrow 5 R2$

$A-A^{T}=\begin{bmatrix} 0 &0 &5 \\ 0 & 0 & -10\\ -5 & 2 & 0 \end{bmatrix}$

$R2 \rightarrow 2R1+R2$

$A-A^{T}=\begin{bmatrix} 0 &0 &5 \\ 0 & 0 & 0\\ -5 & 2 & 0 \end{bmatrix}$

 

Rank of $A-A^{T}=2$

Answer 2

Answer is 2.

Answer 3

A=[2   2   4                 A^T=[2  3 -1       (A-A^T)=[0  0  5                                    (A-A^T)=[0  0  5

      3  -1   2                         3 -1  4                       0  0 -2 (R2=R2+4R1/10)                    0  0  0

     -1   4   5]                       4  2  5]                     -5  2  0]                                                  -5  2  0]

Clearly rank of (A-A^T)==2

Comments

sir, dont't you think that if you find the deteminant of your (A-A^T) which is calculated by you will be '0'  , even for 2 X 2 of your matrix will be zero , so, how can it rank will be '2'

Answer 4

Since determinant of A-A^T = 0 ie scaler 

then rank == (order of the matrix - 1)