in Linear Algebra edited by
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31 votes
Let $P = \begin{bmatrix}1 & 1 & -1 \\2 & -3 & 4 \\3 & -2 & 3\end{bmatrix}$  and $Q = \begin{bmatrix}-1 & -2 &-1 \\6 & 12 & 6 \\5 & 10 & 5\end{bmatrix}$ be two matrices.

Then the rank of $ P+Q$ is ___________ .
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1st column of P+Q is linear combination of 2nd and 3rd column as col1=2*col2-col3. So rank cannot be 3

3
3

Anyone who tried to solve using

Rank(A+B) = Rank(A)+Rank(B) and got answer as 3, the problem is it is not an equality equation.

The equation is:Rank(A+B)  ≤  Rank(A)+Rank(B) 

and 2 is clearly less than 3

2
2

2 Answers

52 votes
52 votes
Best answer
$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$

$\det(P+Q) = 0$, So Rank cannot be $3$, but there exists a $2*2$ submatrix such that determinant of submatrix is not $0$.

So, $\text{Rank}(P+Q) = 2$
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4 Comments

1st and 2nd 

 

or

 

3rd and 2nd

 

0
0

Rank of matrix tells about a lot of things :-

  1. Number of linear independent rows . 
  2. Number of linear independent columns.
  3. Number of non zero rows

One thing to note is that 

number of linear independent rows are equal to number of linear independent columns.

2
2
Rank also tells the number of non zero eigen values right?
0
0
7 votes
7 votes
$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 9 & 10 \\8 & 8 & 8\end{bmatrix}$

Applying $R_2 \leftarrow R_2+R_1$, we get

$P +Q = \begin{bmatrix}0 & -1 &-2 \\8 & 8 & 8 \\8 & 8 & 8\end{bmatrix}$

Since there are only $2$ independent rows $\implies Rank(P+Q) = 2$
Answer:

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