Thread: linear dependence

first row reduce and switch the rows a bit to yield S= [1,0,0,0],[0,a,0,0],[0,0,1,0],[0,1,0,a-1]. I presume you already know that if the rows of S are linearly dependent, then that must mean one of the rows is a linear combination (or scalar multiple) of the others. It becomes apparent that S is linearly dependent when a=0 or a=1. When a=0 the 2nd row is all zeroes, and a row of all zeroes is just a linear combination of the other rows multiplied by zero. When a=1 the 4th and 2nd row are scalar multiples of each other. (4th row times a = 2nd row)

Thank you for your help....but linear dependacy is quite new for me. Is it possible for you to solve this question for me and send me on my email id...sumr_ksa@hotmail.com....I shall be very thankful to you..

1 2 2 a
0 0 a 1
0 1 3 2
0 0 0 a-2

The matrix S is in this form this is a matrix....

I made a small mistake in the previous question:

(4th row times a = 2nd row)

by substituting a for 1, row 4 = row 2 = [0,1,0,0]. technically they are still scalar multiple of each other (multiply by 1).


Sorry but I'm having trouble solving that other matrix. I'm too short on time to devote too much energy on a single question. It's been a while since I've done questions like this. Btw do you know if it's ok to divide a row by a? As far as I know dividing by a adds the restriction that a=/=0, and by introducing a restriction you are changing the solution of the matrix, however in a situation where you have [0,0,0,a|0] (an augmented matrix), it's ok to divide by a since a=/=0 in the first place. I'm not sure if this rule applies more generally to matrices like this though?