One semester, a student gave the following argument to show that if the columns

Question

One semester, a student gave the following argument to show that if the columns of a 2x2 matrix A = [a b c d] are linearly dependent, then det(A) = 0. "If the columns vectors of a 2x2 are linearly dependent, then the two columns are scalar multiples of each other; that is, there is a scalar x such that [a c] = x[b d]. Thus, A = [a b c d] = [xb b xd d]. So, if we take the determinant of A: det [a b c d] = det [xb b xd d] = (xb)d - b(xd) = xbd - bxd = 0. Thus det(A) = 0." Do you find this argument convincing? Why or why not? Write at least 3 sentences.

Explanation / Answer

The argument is correct ,

For ,

Suppose we have two linearly dependent vectors v1 and v2 and let c1 ,c2 be two scalars

Such that c1v1+c2 v2= 0

Then c1 ,c2 both are not zero as v1 ,v2 is linearly dependent ( use Def of linearly dependent )

Suppose c1 not equal zero

c1v1=-c2v2

Divide above equation by c1 as c1 not zero we can divide by c1

v1=(-c2/c1)v2

v1=xv2, x = -c2/c1

Then the matrix A=[ v1 v2]= [xv2 v2]=x[v2 v2]

Det A=x.0=0

as [v2 v 2] have two same column .

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