2.1:10,17,23,32; 2.2:3,6,18,35; 2.3:5,6,8,20
DUE: Friday, Oct. 6, Beginning of Class
2.1:10)
AB = [ 3 -4 ] [ 7 4 ] = [ 1 12-4k ]
[-5 1 ] [ 5 k ] [ -30 -20+k ]
BA = [ 7 4 ] [ 3 -4 ] = [ 1 -24 ]
[ 5 k ] [-5 1 ] [15-5k -20+k ]
Comparing entries, we get AB = BA only if
12-4k = -24 and 15-5k = -30,
which is satisfied for k = 9.
2.1:17) First note that A is 2x2 and AB is 2x3, so B must be 2x3.
B = [ b_1 b_2 b_3 ]. We are asked to find the column vectors b_1 and b_2.
By definition of matrix multiplication, we have
A b_1 = [-1 ] so b_1 is the solution of the system whose
[ 6 ] augmented matrix is below
[ 1 -2 -1 ] [ 1 -2 -1 ] [ 1 0 7 ] so b_1 = [ 7 ]
[-2 5 6 ] ~ [ 0 1 4 ] ~ [ 0 1 4 ] [ 4 ]
Likewise, A b_2 = [ 2 ] gives the system
[-9 ]
[ 1 -2 2 ] [ 1 -2 2 ] [ 1 0 -8 ] so b_2 = [-8 ]
[-2 5 -9 ] ~ [ 0 1 -5 ] ~ [ 0 1 -5 ] [-5 ]
2.1:23) We are given CA = I (I is nxn identity matrix). Note if A is mxn,
then C must be nxm, and below x must be a nx1 vector (that is, x is in R^n).
Show if Ax = 0 (the mx1 zero vector), then x = 0.
Ax = 0 (Multiply both sides on left by C)
--> C(Ax) = C0 = 0 (here 0 is the nx1 zero vector).
Associativity of matrix multiplication gives C(Ax) = (CA)x, so
(CA)x = 0
--> I x = 0
--> x = 0.
2.1:32) (ABx)^T = x^T B^T A^T (where ^T means transpose).
Apply Theorem 3(d) twice to see this.
2.2:3) A = [ a b ] = [ 3 -7 ]
[ c d ] [-6 13 ]
First note ad - bc = 3(13) - (-7)(-6) = -3 does not equal 0, so A is invertible
By Thm.4 in text A inverse = A^-1 = (-1/3) [ 13 7 ]
[ 6 3 ]
2.2:6) The system is Ax = b = [-4 ] , where A is same matrix as in 2.2:3.
[ 1 ]
A is invertible by the previous problem so we have a unique solution
x = (A^-1)b = (-1/3) [ 13 7 ] [-4 ] = (-1/3) [-45 ] = [ 15 ]
[ 6 3 ] [ 1 ] [-21 ] [ 7 ]
2.2:18) A = PB(P^-1) . Note B must be square (same size as P), so A is also
square and of same size.
Multiply both sides of equality on left by P^-1 (inverse) and on right by P:
A = PBP^-1
--> (P^-1)AP = (P^-1)(PBP^-1)P
= ((P^-1)P) B ((P^-1)P)
= IBI
= B
2.2:35) We could perform row operations on [ A I ] to get [ I (A^-1) ],
but then we would be finding all 3 columns of (A^-1).
Since we only need the third column of A^-1, denote this column vector by
x = [ x_1 ] and more efficiently solve the system Ax = [ 0 ] (the third
[ x_2 ] [ 0 ] column of
[ x_3 ] [ 1 ] matrix I)
[ -1 -5 -7 0 ] [ -1 -5 -7 0 ]
[ 2 5 6 0 ] ~ [ 0 -5 -8 0 ] ~
[ 1 3 4 1 ] [ 0 -2 -3 1 ]
[ 1 5 7 0 ] [ 1 0 -1/2 5/2 ]
[ 0 1 3/2 -1/2 ] ~ [ 0 1 3/2 -1/2 ] ~
[ 0 -5 -8 0 ] [ 0 0 -1/2 -5/2 ]
[ 1 0 0 5 ] Thus the 3rd column of A^-1 is [ 5 ]
[ 0 1 0 -8 ] [-8 ]
[ 0 0 1 5 ] [ 5 ]
2.3:5)
[ 5 -9 3 ] [ 1 0 3 ] [ 1 0 3 ] [ 1 0 3 ]
[ 0 3 4 ] ~ [ 0 3 4 ] ~ [ 0 3 4 ] ~ [ 0 3 4 ]
[ 1 0 3 ] [ 5 -9 3 ] [ 0 -9 -12] [ 0 0 0 ]
Only 2 pivot positions so the matrix is not invertible.
2.3:6) Easiest to row reduce to an echelon form.
[ 1 4 2 ] [ 1 4 2 ] [ 1 4 2 ]
[ 2 7 3 ] ~ [ 0 -1 -1 ] ~ [ 0 1 1 ]
[ 1 7 5 ] [ 0 3 3 ] [ 0 0 0 ]
Only 2 pivot positions so not invertible.
2.3:8) Yes, invertible since 4 pivot positions for the 4x4 square matrix.
2.3:20) No, Q can not be invertible. Apply the Invertible Matrix Thm.
Statement (h) false implies statement (a) is false.