Question 1195865: Answer each of the following as True or False justifying your answers:
If S and K are n×n matrices such that S is symmetric and K is skew symmetric, then |S+K|=|S-K|. If |■(a&a&a-1&c@a&a&a-1&b@a&a&a+1&b@a&a+1&b&c)|=0, then a=b=c=0. If A,B and C are non-singular n×n matrices such that AB=C, BC=A and CA=B, then |ABC|=1. If A and B are 3×3 matrices, then AB-AB^T is a non-singular matrix. The vector U=(2,-1,-1) in R^3 is a linear combination of the vectors V_1=(1,2,7),V_2=(2,5,17) and V_3=(-1,2,5).
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
Answer each of the following as True or False justifying your answers:
(a) If S and K are n×n matrices such that S is symmetric and K is skew symmetric,
then |S+K|=|S-K|.
(b) If |■(a&a&a-1&c@a&a&a-1&b@a&a&a+1&b@a&a+1&b&c)|=0, then a=b=c=0.
(c) If A,B and C are non-singular n×n matrices such that AB=C, BC=A and CA=B, then |ABC|=1.
(d) If A and B are 3×3 matrices, then AB-AB^T is a non-singular matrix.
(e) The vector U=(2,-1,-1) in R^3 is a linear combination of the vectors
V_1=(1,2,7), V_2=(2,5,17) and V_3=(-1,2,5).
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(c) We are given for non-singular nxn-matrices A, B and C
AB = C, BC = A, CA = B.
Multiply all three equations (both sides). You will get then
A*B*B*C*C*A = A*B*C.
It implies for determinants
|A|*|B|*|B|*|C|*|C|*|A| = |A|*|B|*|C|, or
|A*B*C|^2 = |A*B*C|.
Since |ABC| =/= 0 (due to non-singularity), we can reduce/cancel common factor |A*B*C| in both sides.
We then get |A*B*C| = 1, QED.
ANSWER. The statement is TRUE.
(d) The statement is FALSE.
To prove that it is false, consider A= 0, B= 0 (zero matrices).
(e) We want to check if there are real numbers "a", "b" and "c" such that
U = a*V_1 + b*V_2 + c*V_3.
It is the same as to check, if this system of three linear equations has a solution
for real numbers "a", "b" and "c"
1*a + 2*b - 1*c = 2, (E1)
2*a + 5*b + 2*c = -1, (E2)
7*a + 17*b + 5*c = -1. (E3)
Notice that for left sides of equations (E1), (E2), (E3) we have their linear combination 1*E1 + 3*E2 = E3.
For right side terms, 1*E1 + 3*E2 = 1*2 + 3*(-1) = 2 - 3 = -1 is the same as E3.
So, we can expect that the system has a solution.
Now, I am laizy to check it directly, and we live in XXI century - so I use an online solver
https://www.emathhelp.net/calculators/linear-algebra/gauss-jordan-elimination-calculator/
It confirms that the system (E1), (E2), (E3) has the solution a= 12, b= -5, c= 0.
ANSWER. Vector U=(2,-1,-1) in R^3 is a linear combination of the vectors
V_1=(1,2,7), V_2=(2,5,17) and V_3=(-1,2,5): U = 12*U_1 - 5*V_2.
Parts (c), (d) and (e) are solved, answered and explained.
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