recategorized by
1,346 views
0 votes
0 votes

Consider the rank of matrix $'A'$ of size $(m \times n)$ is $"m-1"$. Then, which of the following is true?

  1. $AA^T$ will be invertible.
  2. $A$ have $"m-1"$ linearly independent rows and $"m-1"$ linearly independent column.
  3. $A$ will have $"m"$ linearly independent rows and $"n"$ linearly independent columns.
  4. $A$ will have $"m-1"$ linearly independent rows and $"n-1"$ independent columns.
recategorized by

3 Answers

1 votes
1 votes

Here it is given rank of matrix $m-1$

That means there are $m-1$ non zero rows in the matrix 

As, if the definition of linearly independent is the determinant of the matrix must be non-zero, otherwise in non linearly independent matrix determinant of the matrix must be 0. 

Now, as the matrix has rank $m-1$ , that means matrix is a square matrix

Because, without square matrix we cannot find determinant

From here we can say, there must be $m-1$ column which are non-zero.

So, option B) is correct

for more information here

0 votes
0 votes

these basics of rank is useful to do such theoritical problems.

0 votes
0 votes
Rank of matrix A=The number of linearly independent rows of A=the number of linearly independent columns of A

As in question rank of matrix is given m-1 so it will have m-1 linearly independent rows and m-1 linearly independent columns

Therefore option B is correct

Related questions

3 votes
3 votes
3 answers
1
DebRC asked Sep 19, 2022
1,014 views
Let $A$ be a $3$ x $3$ matrix with rank $2$. Then, $AX=0$ hasThe trivial solution $X=0$.One independent solution.Two independent solution.Three independent solution.
0 votes
0 votes
0 answers
3
Mohib asked Oct 17, 2022
451 views
While studying Linear algebra I got 2 perspectives. Which meaning out of these 2 is more accurate?
0 votes
0 votes
2 answers
4
Overflow04 asked Sep 21, 2022
539 views
If A is a non-zero column matrix of order n×1 and B is a non-zero row matrix of order 1×n then rank of AB equals ? Rank(ab) can be zero???