The matrix will only be consistent if the last entry of row three is 0, meaning 2b 3 + b 2 b 1 = 0.

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1 ) Find all solutions of the linear system. Express the answer in vector form. x + 2x + x + x 5 = 2 2x 2 + 2x + 2x + x 5 = 8 x + 2x + x + 9x 5 = 2 2 Solution: Reduce the augmented matrix [ ] to get [ ]. This can be interpreted to mean 5 x + 2x x 5 = ; x 2 + x x 5 = ; x + 5x 5 =, or x = 2x + x 5 x 2 = x + x 5 x = 5x 5 In vector form, this is x x 2 x = x = 2) Consider the system x [ x 5 ] 2x + x 5 x + x 5 x = 5x 5 [ x 5 ] [ ] x + 7x 2 x = b x + x 2 x = b 2 x + 2x 2 = b + x 2 + x 5. 5 [ ] [ ] a) Give conditions on the numbers b, b 2, b so that the system is consistent. 7 b Solution: Reduce the matrix [ b 2 ] : 2 b 7 b 7 b [ b 2 ] R 2 R [ 2 2 b 2 b 2 b R R 2 b b 2 ] 7 b [ 2 2 b 2 b ] 2R + R 2 2b + b 2 b The matrix will only be consistent if the last entry of row three is, meaning 2b + b 2 b =. b b) Give an example of a vector [ b 2 ] such that the linear system is inconsistent. b Solution: Any vector where 2b + b 2 b will make the system inconsistent. For example, b b = [ b 2 ] = [ ]. b

2 ) Show that the vectors 2 v = [ 2 ], v 2 = [ ], v = [ ], v = [ ] are linearly dependent, and find a linear combination of the vectors that equals θ and has nonzero coefficients. Solution: Since there are four vectors and each vector is an element of R, the set of vectors must be dependent. To find a linear combination that s equal to θ, we must solve the equation x v + x 2 v 2 + x v + x v = θ. This is equivalent to solving Ax = θ, where A = [v, v 2, v, v ]. 2 This equation can be solved by reducing the matrix [ 2 ]. Reducing the matrix 2 gives [ ]. This can be interpreted to mean x = 2 5 x, x 2 = 5 x, and x = 6 5 x. Letting x = gives one possible solution: 2 5 v + 5 v v + v = θ. 5 ) Is the matrix A = [ 2 ] singular or nonsingular? If it is nonsingular, find its inverse matrix. 2 Solution: The matrix is nonsingular because it is row-equivalent to I (performing Gauss-Jordan elimination on A results in the matrix I). We find A by performing Gauss-Jordan elimination on the augmented matrix [A I]: 5 5 [ 2 ] [ 2 ] 2 R R 7 2R 7R 2 5 R R 2 2R + R 2 [ 2 ] R 2 + R [ ] R 6 2 [ ] R 2 2 [ ] R The three columns on the right of the reduced matrix form A, so 2 A = [ ] You can verify this is correct by showing that the product AA = I = A A.

3 5) a) Suppose A and B are 2 2 matrices that satisfy Find A and B. (2A + B) T = [ 5 2 ], A + B = [ 6 ]. Solution: Take the transpose of the first equation to get [(2A + B) T ] T = 2A + B = [ ]. Subtract the second equation from this equation to get A = [ ] [ ] = [2 2 ]. Then by the second equation, B = [ ] A = [ ] [2 2 ] = [ 2 ]. b) Suppose C is a 2 2 matrix that satisfies ((C) ) T = [ 5 2 ]. Find C. Solution: Take the transpose of both sides: (C) = [ 5 ], then take the inverse of both 2 sides: C = [ 5 2 ] = [ 2 5 ] = [ ]. Finally, divide both sides by to get 2 5 C = [ 2 5 ] 6) Suppose A and B are nonsingular matrices such that A = [ 7 2 ], B = [ 7 5] 2 a) Without calculating A or B, find (A). Solution: (A) = A = [ ] b) Without calculating A or B, find (A T B). Solution: (A T B) = B (A T ) = B (A ) T = [ 7 5] [ 2 ] = [ ]

4 c) Without calculating A or B, find [(A B ) A B]. Solution: [(A B ) A B] = [((B ) (A ) )A B] = [(BA)A B] = [B(AA )B] = [BIB] = [BB] = B B = [ 7 5] [ 7 5] = [ 9 2] ) Suppose A is an n n matrix such that A A 2 2A + 8I = O. Show that A is nonsingular and A = 8 (2I + A A2 ). Solution: Rearrange the given equation: A(A 2 A 2I) = 8I, then divide both sides by 8 to get A [ 8 (A2 A 2I)] = I. This simultaneously shows that A is nonsingular and that A = 8 (A2 A 2I). (A is nonsingular because multiplying it by another matrix produces the nonsingular matrix I, and according to the lemma we proved in class, a nonsingular matrix can only result from the product of two nonsingular matrices. It shows that A = 8 (A2 A 2I) since A multiplied by 8 (A2 A 2I) results in the identity matrix I.) 8) Suppose the v, v 2, and v are linearly independent vectors in R n. Show that the vectors w = v, w 2 = 2v + v 2, and w = v + 2v 2 + v are also linearly independent. Solution: We would like to show that w, w 2, and w are linearly independent. That means we need to show that the equation x w + x 2 w 2 + x w = θ has only the trivial solution (i.e. x = x 2 = x = ). Substituting gives x v + x 2 (2v + v 2 ) + x (v + 2v 2 + v ) = θ. Rearranging terms in this equation gives (x + 2x 2 + x )v + (x 2 + 2x )v 2 + x v = θ. Since we know that v, v 2, and v are linearly independent, we know that the three constants in this equation must equal to zero, thus we have x + 2x 2 + x = x 2 + 2x = x = This means that x = x 2 = x =, so therefore w, w 2, and w are linearly independent.

5 9) Consider the matrix A = [ x x2 ], where x is any real number. x a) Show that A 2 = I for any real x. Solution: Multiplying A by itself gives [ x x2 ] [x x2 x x ] = + x 2 x( x 2 ) x( x 2 ) [x2 x x x 2 + x 2 ] = [ ] = I b) Show that A is nonsingular for any x. Solution: To be nonsingular, we need Δ. For matrix A, Δ = x 2 ( x 2 ) = So Δ = for all x. c) Find all values of x for which A is a symmetric matrix. Solution: To be symmetric, we need A T = A. For this matrix, we need x [ x 2 ] = [x x2 x x ]. So, to be symmetric, we need x 2 =, so the only way that A will be symmetric is if x =. ) Let I be the n n identity matrix and v a vector in R n of length. Show that the matrix A = I vv T satisfies the identity A 2 = A. Solution: Since v has length, that means v =, which means v T v = (since v T v = ). Now, A 2 = AA = (I vv T )(I vv T ) = I 2 I(vv T ) (vv T )I + (vv T )(vv T ) = I 2vv T + v(v T v)v T = I 2vv T + vv T = I vv T = A.

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