# Thread: linear algebra urgent help please

1. ## linear algebra urgent help please

true or false and why?
1) every square matrix can be written as the product of elementary matrices.
2) The inverse of an elementary matrix is an elementary matrix.
3) Every system with 3 equations in 4 unknowns has at least one solution.

4) A square matrix is defined to be stochastic if all its elements are nonnegative and the sum of the elements in any given column is 1.
Prove that the product of any two stochastic matrices is stochastic.

please help me urgently. Thank you very much.

2. Originally Posted by happystudent
true or false and why?
1) every square matrix can be written as the product of elementary matrices. Mr F says: False since the statement is not true when det(M) = 0.

2) The inverse of an elementary matrix is an elementary matrix. Mr F says:
True.

3) Every system with 3 equations in 4 unknowns has at least one solution. Mr F says:
False. It's not hard to come up with a counter-example to the statement.

4) A square matrix is defined to be stochastic if all its elements are nonnegative and the sum of the elements in any given column is 1.
Prove that the product of any two stochastic matrices is stochastic.
[snip]
Where are you stuck with 4)?

3. i don't understand 4. so i don't know what to start

4. Originally Posted by happystudent
[snip]
4) A square matrix is defined to be stochastic if all its elements are nonnegative and the sum of the elements in any given column is 1.
Prove that the product of any two stochastic matrices is stochastic.
[snip]
Originally Posted by mr fantastic
Where are you stuck with 4)?
Originally Posted by happystudent
i don't understand 4. so i don't know what to start
Let C = AB where A and B are both nxn stochastic matrices (column).

By definition:

Sum of entries in rth column of A: $\displaystyle \sum_{i=1}^{n} a_{ir} = 1$

Sum of entries in rth column of B: $\displaystyle \sum_{i=1}^{n} b_{ir} = 1$

By definition of matrix multiplication:

Element of C in ith row and jth column: $\displaystyle c_{ij} = \sum_{r=1}^{n} a_{ir} b_{rj}$.

Therefore the sum of the entries in the jth column of C is $\displaystyle \sum_{i=1}^{n} c_{ij} = \sum_{i=1}^{n} \sum_{r=1}^{n} a_{ir} b_{rj}$

$\displaystyle = \sum_{r=1}^{n} \left( \sum_{i=1}^{n} a_{ir} \right) b_{rj} = \sum_{r=1}^{n} (1) b_{rj} = (1) (1) = 1$

where the justification of the last line is left for you.