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Solution to Linear Algebra Hoffman & Kunze Chapter 1.5


Exercise 1.5.1

We have AB=[44],so
ABC=[44][1  1]=[4444].and
CBA=[1  1][44]=[0].


Exercise 1.5.2

We haveA2=[111201301][111201301]=[211523634].And
AB=[111201301][221344]=[5180102].Thus
A2B=[211523634][221344](1)=[73204255].And
A(AB)=[111201301][5180102](2)=[73204255].Comparing (1) and (2) we see both calculations result in the same matrix.


Exercise 1.5.3

For example, [0100],[0010]are two such matrices.


Exercise 1.5.4

A=[111201301]E1A=[100210001][111201301]=[111021301]E2(E1A)=[100010301][111021301]=[111021030]E3(E2E1A)=[10001/20001][111021030]=[111011/2030]E4(E3E2E1A)=[110010001][111011/2030]=[101/2011/2030]E5(E4E3E2E1A)=[100010031][101/2011/2030]=[101/2011/2003/2]E6(E5E4E3E2E1A)=[100010002/3][101/2011/2003/2]=[101/2011/2001]E7(E6E5E4E3E2E1A)=[101/2010001][101/2011/2001]=[100011/2001]E8(E7E6E5E4E3E2E1A)=[100011/2001][100011/2001]=[100010001].


Exercise 1.5.5

To find such a C=[abcdef] we must solve the equation
[abcdef][112210]=[3144].This gives a system of equations
a+2b+c=3a+2b=1d+2e+f=4d+2e=4We row-reduce the augmented coefficient matrix
[1210003120000100012140001204][101/20001011/40001000101/24000011/40].Setting c=f=4 gives the solution
C=[104614].Checking:
[104614][112210]=[3144].


Exercise 1.5.6

The ij-th entry of AB is r=1kAirBrj. Since the term Brj is independent of i, we can view the sum independent of i as r=1nBrjαr where αr is the r-th column of A. I’m not sure what more to say, this is pretty immediately obvious from the definition of matrix multiplication.


Exercise 1.5.7

Suppose
A=[abcd],B=[xyzw].AB=[ax+bzay+bwcx+dzcy+dw].Then AB=I implies the following system in u,r,s,t has a solution
au+bs=1cu+ds=0ar+bt=0cr+dt=1because (x,y,z,w) is one such solution. The augmented coefficient matrix of this system is
(3)[a0b01c0d000a0b00c0d1].As long as adbc0 this system row-reduces to the following row-reduced echelon form
[1000d/(adbc)0100b/(adbc)0010c/(adbc)0001a/(adbc)]Thus we see that necessarily x=d/(adbc), y=b/(adbc), z=c/(adbc), w=a/(adbc). Thus
B=[d/(adbc)b/(adbc)c/(adbc)a/(adbc)].Now it’s a simple matter to check that
[d/(adbc)b/(adbc)c/(adbc)a/(adbc)][abcd]=[1001].

The loose end is that we assumed adbc0. To tie up this loose end we must show that if AB=I then necessarily adbc0. Suppose that adbc=0. We will show there is no solution to (3), which contradicts the fact that (x,y,z,w) is a solution. If a=b=c=d=0 then obviously ABI. So suppose WOLOG that a0 (because by elementary row operations we can move any of the four elements to be the top left entry). Subtracting ca times the 3rd row from the 4th row of (3) gives
[a0b01c0d000a0b00ccaa0dcab1].Now ccaa=0 and since adbc=0 also dcab=0. Thus we get
[a0b01c0d000a0b000001].and it follows that (3) has no solution.


Exercise 1.5.8

We want to know when we can solve for a,b,c,d,x,y,z,w such that
[c11c12c21c22]=[abcd][xyzw][abcd][xyzw]The right hand side is equal to
[bzcyay+bwbxdycx+dzazcwcybz]Thus the question is equivalent to asking: when can we choose a,b,c,d so that the following system has a solution for x,y,z,w
(4)bzcy=c11ay+bwbxdy=c12cx+dzazcw=c21cybz=c22The augmented coefficient matrix for this system is
[0cb0c11bad0bc12c0dacc210cb0c22]This matrix is row-equivalent to
[0cb0c11bad0bc12c0dacc210000c11+c22]from which we see that necessarily c11+c22=0.

Suppose conversely that c11+c22=0. We want to show A,B such that C=ABBA.

We first handle the case when c11=0. We know c11+c22=0 so also c22=0. So C is in the form
[0c12c210].In this case letA=[0c12c210],B=[1000].ThenABBA=[0c12c210][1000][1000][0c12c210]=[00c210][0c1200]=[0c12c210]=C.So we can assume going forward that c110. We want to show the system (4) can be solved. In other words we have to find a,b,c,d such that the system has a solution in x,y,z,w. If we assume b0 and c0 then this matrix row-reduces to the following row-reduced echelon form
[10(da)/c1dabcc11c12b01d/c0c11/c0000c11b(da)+c21+c12cb0000c11+c22]We see that necessarily
c11b(da)+c21+c12cb=0.Since c110, we can set a=0, b=c=1 and d=c12+c21c11. Then the system can be solved and we get a solution for any choice of z and w. Setting z=w=0 we get x=c21 and y=c11.

Summarizing, if c110 then:
a=0,b=1,c=1d=(c12+c21)/c11x=c21,y=c11,z=0,w=0For example if C=[2132] then A=[0112] and B=[3200]. Checking:
[0112][3200][3200][0112]=[2132].

From http://greggrant.org

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
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