If X_1 and X_2 are solutions of the equation AX = B (B notequal 0), then X_1 + X

Question

If X_1 and X_2 are solutions of the equation AX = B (B notequal 0), then X_1 + X_2 is also a solution. If A and B are 2 times 2 matrices, then the sum of the terms on the main diagonal of AB - BA is zero. Suppose A is a 2 times 2 matrix. If A is invertible, then A^t is also invertible. Suppose that S = {v_1 v_2. V_3} f is a linearly independent set in R^n; then T = {v_1, v_2, v_1 + v_2 + v_3} is also linearly independent. For a elementof R, the vectors [a^2 0 1] [0 a 2] and [1 0 1] are linearly independent.

Explanation / Answer

2. Let A = [ a b ]   and B = [ e f ] Then AB = [ ae+bg   af+bh ]                 BA = [ ae+cf    be+df ]

[ c d ] [ g h ] [ ce+dg   cf+dh ]                          [ag+ch   bg+dh ]

Then AB –BA = [ bg-cf                   af+bh-be-df ]

[ ce+dg –ag-ch            cf-bg    ]

so that the sum of the terms on the main diagonal of AB – BA = bg –cf +cf –bg = 0. Hence the statement is true.

3. Let A = [ a b ] . Then AT = [ a c ] . Thus det(A) = ad-bc and det(AT) = ad-bc. Now, if A is

[ c d ] [ b d ]

invertible, then det(A) 0 so that det(AT) 0. Hence AT is invertible. The statement is true.

4. Let the vectors v1, v2, v1+v2+v3 be linearly dependent. Then there exist scalars a,b, c , not all 0 such that av1 +bv2 + c(v1 +v2 +v3) = 0. Then ( a+c)v1 + (b+c)v2+cv3 = 0. Now, since v1 , v2, v3 are linearly independent, hence a=c = 0, b+c = 0 and c = 0 so that a= 0 and b = 0. This is a contradiction. Hence the vectors v1, v2, v1+v2+v3 are linearly independent.

5. Let the vectors (a2,0,1)T , (0,a,2)T and (1,0,1)T be linearly dependent. Then there exist scalars p,q,r, not all 0 such that p(a2,0,1)T + q(0,a,2)T + r (1,0,1)T = 0. Then pa2+ r = 0, qa = 0 and p+2q +r = 0. If a 0, then q = 0 (as qa = 0). Further, then p+r = 0 or, r = -p so that pa2 –p = 0 or p(a2 -1) = 0. Now, if a ± 1, then p = 0 so that r = 0. This is a contraction. Hence the vectors (a2,0,1)T , (0,a,2)T and (1,0,1)T are linearly independent if a 0 and a ± 1. Thus, the statement is true if a 0 and a ± 1.

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