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Problems #13,24 on pg.229 are graded.
#13,pg.229: Determine the dimensions of NulA and ColA for the matrix
A=[1−690−2012−450005100000].
Solution: This matrix is in echelon form. If we compute the reduced echelon form we get
[1−690−2012−450005100000]∼[102101645012029500011500000].
Notice that this is the coefficient matrix of the matrix equation A→x=→0 (notice that →x must be a 5×1 vector; the augmented matrix adds a column of zeros on the right side of this matrix) and so the matrix equation A→x=→0 yields the following system of equations:
{x1+21x3+1645x5=0x2+2x3+295x5=0x4+15x5=0
whose solution is therefore
→x=x3[−212100]+x5[16452950−151].
We see that there are two free variables in the solution of A→x=→0 and therefore dimNulA=2. From the original matrix A we see that there are three pivot columns, therefore dimColA=3.
#24, pg.229: The first three Laguerre polynomials are L0(x)=1,L1(x)=−t+1, and L2(x)=t2−4t+2. It is known that B={L0,L1,L2} forms a basis of P2. Find the coordinate vector of the vector →p(t)=−2t2+5t+5 with respect to B, taking b0=L0,b1=L1,b2=L2.
Solution: Our goal here is to compute the column vector [→p]B. This is done by finding the weights c0,…,c3 such that
c0L0+c1L1+c2L2=→p,
or in other words
c0+c1(−t+1)+c2(t2−4t+2)=−2t2+5t+5.
Simplify the left hand side to get
t2c2+t(−c1−4c2)+(c0+c1+2c2)=−2t2+5t+5.
Equating coefficients yields
{c2=−2−c1−4c2=5c0+c1+2c2=5
yielding
{c2=−2c1=3c0=6.
Hence
[→p]B=[−236].
