Linear Algebra-Linear dependence

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The discussion focuses on proving the linear dependence of the columns of a 3x3 matrix U under specific conditions where a=0, d=0, or f=0. It is established that if either a=0 or d=0, the columns can be shown to be dependent through scalar multiplication leading to identical vectors. The challenge arises in proving the case where f=0, which results in the vectors (a,0,0), (b,d,0), and (c,e,0). The solution involves finding scalars x and y to express one vector as a linear combination of the others, leading to two equations that can be solved based on the value of d. The discussion concludes that if d=0, there are infinitely many solutions, confirming the linear dependence of the columns.
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Homework Statement



let U be a 3x3 matrix containng columns C1, C2, C3. The three column vectors C1= (a,0,0) , C2=(b,d,0), C3=(c,e,f)
prove that if a=0 or d=0 or f=0 (3cases), the columns of U are dependent?
problem from Linear algebra and applications, fourth editon, Gilbert strang


Homework Equations


no eqations


The Attempt at a Solution


I successfully proved the first two cases
if a=0, if we multiply C3 or C2 with zero then C1 will be equal to C2 or C3. The columns become independent
if d=0, if we multiply C1 with b/a and C2 with a/b, C1 and C2 will be identical and the columns become independent.
But i don't know how to prove the third case, i tried with different comibinations of scalars with multiply with C2 and C3, but i can't make this two columns identical. Enlighten me.
 
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The third case has f= 0 so the three vectors are (a,0,0), (b, d, 0), and (c, e, 0).
You want to find numbers x and y such that (c, e, 0)= x(a, 0, 0)+ y(b, d, 0).
That is equivalent to the two equations c= xa+ yb and e= yd. Obviously, if d is not 0, y= e/d and your first equation become c= xa+ be/d. It should be easy to solve that for x. If d= 0 there are an infinite number of soutions.
 

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