..READ THIS
FIRST..Problems
about Matrices, Determinants, Systems of linear equations
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about Matrices, Determinants, Systems of linear equations
Level 2 problems
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A and B are n x n matrices. Investigate if (A + B)2 = A2 + 2.A.B + B2(A + B)2 = (A + B).(A + B) = A.A + A.B + B.A + B.B = A2 + A.B + B.A + B2 Since the product is not commutative, we do NOT have (A + B)2 = A2 + 2.A.B + B2 -
We say that two matrices A and B commute, if and only if AB = BA. Show that all matrices like [a -b] [b a] commute.[a -b] [a' -b'] = [aa'-bb' -ab'-ba'] [b a] [b' a'] [ba'+ab' -bb'+aa'] and [a' -b'] [a -b] = [aa'-bb' -ab'-ba'] [b' a'] [b a] [ba'+ab' -bb'+aa'] -
Prove that for each n x n matrix A , AT .A = symmetric
We'll prove that AT .A is equal to its transpose. (AT .A)T = AT .(AT)T = AT .A
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Prove that if A is skew-symmetric, then A.A is symmetric.
A is skew-symmetric <=> A = - AT . So, (A.A)T = AT .AT = (- A ).(- A ) = A.A Hence, A.A is symmetric.
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Given X = [x y] and A is a 2 x 2 matrix . All elements of the matrices are real numbers. Examine if X . AT . A . XT can be negative.
First we see that X . AT . A . XT is a 1 x 1 matrix. It is a number, so maybe it is negative. But, X . AT = is a 1 x 2 matrix, say [a b]. A . XT = (X . AT )T = [a b]T Then X . AT . A . XT = (X . AT ).(X . AT )T = [a b] .[a b]T = a2 + b2 So, X . AT . A . XT can not be negative. -
[ 1 m - 1 2 m - 3 ] Given A = [ m 2 m - 2 2 ] [ m + 1 3 m - 3 m.m - 1 ] Calculate the condition for m such that A is regular. Assume that m satisfies this condition and consider the system with A as coefficient matrix. / x + (m - 1)y + (2 m - 3)z = 1 | | m x + (2 m - 2)y + 2 z = 0 | \ (m + 1)x + (3 m - 3)y + (m.m - 1)z = 0 Now, let x = 1 and calculate the values of m such that the system has a solution for y and z.2 Det(A) = m (1 - m) (m - 2) A is regular if and only if m is NOT in { 0, 1, 2 } With x = 1, the system becomes / (m - 1)y + (2 m - 3)z = 0 | | (2 m - 2)y + 2 z = - m | \ (3 m - 3)y + (m.m - 1)z = -(m + 1) The coefficient matrix of the first and the second equation is [m - 1 2 m - 3 ] [2 m - 2 2 ]The determinant is -4 (m - 1) (m - 2) and since m is not in { 0, 1, 2 }, the determinant is not 0.
So, we consider the first and the second equation as main equations and the system has a solution for y and z if and only if the characteristic determinant of the third equation is 0.The condition is |m - 1 2 m - 3 0 | |2 m - 2 2 - m | = 0 |3 m - 3 m.m - 1 - m - 1| <=> 2 (m + 4) (m - 1) (m - 2) = 0Since m is not in { 0, 1, 2 }, there are no real values of m such that the system has a solution for y and z. -
The matrix x is a 2 x 2 matrix. Calculate three solutions of the quadratic equation x2 - x = 0x2 - x = 0 <=> x (x-I) = 0x = 0-matrix and x = Identity matrix are two solutions, but we know that x (x-I) can be zero without x = 0 or x = I.To find another solution of the given equation we'll search for a matrix
[ 1 b ] x = [ ] such that x2 = x [ c d ] x2 = x <=> [ 1 b ] [ 1 b ] [ 1 b ] [ ].[ ] = [ ] [ c d ] [ c d ] [ c d ] <=> [ 1 + b c b + b d ] [ 1 b ] [ ] = [ ] [ c + d c c b + d2 ] [ c d ] <=> [ b c b d ] [0 0 ] [ ] = [ ] [ c d b c + d2 - d ] [0 0 ]Since we need only one solution, we choose d = 0 and c = 0.
Then b = an arbitrary number. So, even with d = 0 and c = 0, we find an infinity number of solutions.[ 1 b ] x = [ ] [ 0 0 ]
Level 3 problems
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[-2 -9 ] n [1-3n -9n ] Given A = [ ] . Prove that A = [ ] [ 1 4 ] [ n 1+3n ]We'll prove this by complete induction. a) The property is trivial for n = 1 b) Assume the property is true for n = k. Then we have to prove that the property is true for n = k+1. k+1 k [1-3k -9k ] [-2 -9 ] [-2-3k -9-9k] A = A .A = [ ].[ ] = [ ] [ k 1+3k ] [ 1 4 ] [k+1 3k+4] [1-3(k+1) -9(k+1) ] = [ ] [ k+1 1+3(k+1) ] -
[1 1 0 ] Given : A = [0 1 0 ] [0 0 1 ] Show that A is regular. n Calculate A . Calculate the real numbers a and b such that A2 + a A + b I = 0 ( I is the 3 x 3 identity matrix) Show that there are real numbers c0,c1,c2, ... ,cn such that A-n = c0.I + c1.A2 + c2.A3 + c3.A4 + c4.A + ... + cn.AnDet(A) = 1 , so A is regular. 2 [1 2 0 ] 3 [1 3 0 ] A = [0 1 0 ] A = [0 1 0 ] [0 0 1 ] [0 0 1 ] From this we'll prove by complete induction that n [1 n 0 ] A = [0 1 0 ] [0 0 1 ] This expression is true for n = 1 Assume the property is true for n = k. Then we have to prove that the property is true for n = k+1. k+1 k [1 k 0 ][1 1 0 ] [ 1, k + 1, 0 ] A = A .A = [0 1 0 ][0 1 0 ] = [ 0, 1, 0 ] [0 0 1 ][0 0 1 ] [ 0, 0, 1 ] A2 + a A + b I = 0 <=> [1 2 0 ] [1 1 0 ] [1 0 0 ] [0 0 0 ] [0 1 0 ] + a [0 1 0 ] + b [0 1 0 ] = [0 0 0 ] [0 0 1 ] [0 0 1 ] [0 0 1 ] [0 0 0 ] <=> 1 + a + b = 0 2 + a = 0 <=> a = -2, b = 1 From this we have A2 - 2 A + I = 0 <=> 2 A - A2 = I <=> A (2 I - A ) = I From this, we see that (2 I - A ) is the inverse of A A-1 = 2 I - A => A-n = (2 I - A )n => A-n = a polynomial in A of degree n, with real coefficients => there are real numbers c0,c1,c2, ... ,cn such that A-n = c0.I + c1.A + c2.A2 + c3.A3 + c4.A4 + ... + cn.An
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