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MATH 244 -- Linear Algebra 

Problem Set 3 Solutions 

February 9, 2007 

1.8/4.  Since ,  this question is asking 

whether the system has any solutions, and 

if so, whether the solution is unique. 

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Since the rref has a pivot in each row of A, there is a unique  x =  

1.8/5  Same idea as number 4 above: 

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The system is consistent, but there is a free variable so the  

solution is not unique.  The general solution looks like 

where is arbitrary. 

1.8/10.  This question asks for all the solutions of the  

homogeneous system   

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Because of the row of zeroes, this system has a free variable (.  The 

solutions can be parametrized as: 

where  is arbitrary. 

1.8/12.  Now we consider the same coefficient matrix as in 10, but  

the question is whether  Ax = b  has solutions for b ≠ 0. 

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The inhomogeneous system is inconsistent (look at the last row). 

Hence  b  is not in the range of this linear transformation. 

1.8/19.  We are given that and   

Hence the standard matrix of T is A =Then,  

If  

1.8/27 

a)  The usual parametrization of the line through  p  and  q  uses 

the direction vector  v = q - p,  and the points on the line  

are  x = p + t(q - p),  for all This can be rearranged 

using the distributive law for scalar multiplication over vector 

sums:  x = (p + tq.    The line segment from p to q  is 

the portion of the line with   

b)  If  T  is a linear transformation then applying T  to the points 

on the line segment from part a and using the definition of  

linearity yields: 

   (p + tq) = T((1 - t)p) + T(tq) 

                           = (1 - t) T(p) + t T(q) 

When we have the points on the line segment  

from T(p)  to  T(q)  (note the final equation here looks 

just like the parametrization of the line segment from  

p to q in the domain of the mapping, but it gives points 

in ) 

If  T(p) ≠ T(q), then we have an actual line segment. 

If T(p) = T(q), though, then we get the same point for  

all the line segment is collapsed to a point by  

T).   

1.8/28  The idea is similar to number 27.  The points in 

the parallelogram spanned by  u, v  are the  au + bv,  

where 0 ≤ a ≤ 1  and If we apply  T  to any 

one of these points, using the properties in the definition 

of linearity, we get 

T( au + bv) = T(au) + T(bv)  

                        = a T(u) + b T(v) 

This gives one of the points in the parallelogram spanned 

by  T(u)  and  T(v)  (if it's actually a parallelogram). Comment: 

As in the previous problem there are also "special cases"  

depending on the mapping and the actual vectors.  In  

some cases, the parallelogram could be mapped onto a  

line segment (if T(u) and T(v)  are collinear), or a single 

point (if T(u) and T(v) are both 0). 

1.9/2  By the definition of the standard matrix of T, we 

put T() in the first column T() in the second  

column and T() in the third: 

1.9/4.  The process is the same is in 2 above, but 

now the matrix will be 2 x 2 since we the domain is  

The vectors  T()  and T()  are found by  

trigonometry -- in an isosceles right triangle with 

hypotenuse of length 1, the legs have length  

Since we rotate by  π/4  clockwise,  T()  is 

in the 4th quadrant (x > 0, y < 0)  and T() is in the  

first quadrant. 

So A =  

1.9/6.  Using the description given in the problem, 

1.9/8.   Applying the two steps making up  T in succession,    

we get 

so  

(Comment:  This is the same transformation 

as a counter-clockwise rotation through an angle of π/2 

about the origin(!))  

1.9/31.  "T is one-to-one if and only if A has 

n  pivot columns."  (Note:  This means every column 

in  A  is a pivot column, because  A  has  n   

columns all together).  

This is true because T  is one-to-one if and only  

if the homogeneous system  A x = 0  has only the  

trivial solution (Theorem 11).   T  is  one-to-one 

if and only if there are no free variables in the  

homogeneous system.  

1.9/32.  "T maps   onto  if and only if  A   

has  m   pivot columns."  This is true because saying 

T  is onto is the same as saying  A x = b  is 

consistent for  all  equivalently,  

that  the columns of A  span This means that 

every row must have a pivot, so there are  m   

columns containing pivots.  (Theorem 12) 

1.9/35.  If  T  is one-to-one then   

If T  is onto, then These assertions 

follow from problems 31 and 32 above, since 

the matrix of T  is  n x m.   

1.9/36.  For all x, y and scalars c  we have 

        T( S ( x + y )) = T ( S (x) + S (y) )   (since S  is linear) 

                             = T ( S (x) ) + T ( S (y) )  (since  T  is linear) 

and   

         T( S ( c x ) ) = T( c S ( x ) )  (since  S  is linear) 

                           = c T( S ( x ) )  (since  T  is linear) 

Since the composite mapping satisfies the two properties 

in the definition, it is also linear. 

2.2/3. 

We can follow the process described in class: 

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Then the inverse is the right-hand 2 x 2 block 

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2.2/6  When  A  is invertible,  A x = b  has a unique 

solution for all  b.  The solution is   x =  

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2.2/15.  The matrix   

2.2/18.  To solve for  B,  we have to realize 

that matrix multiplication is not commutative.  This means, 

for instance, that if we multiply both sides of the equation 

by P  to cancel the    on the right of  B,  then that P 

must go on the right on both sides of the equation.   

Hence  

2.2/19.  Proceed as in 18: 

Multiply both sides of the equation by  and  

(and group terms by associativity) 

2.2/31 

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2.2/32 

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From these examples, you should conjecture that the  

inverse of the general matrix  A  of this form is the  

matrix  B  with 1 on the main diagonal,  -1  on 

the subdiagonal, and zeroes everywhere else.     

There are many ways to show  that  AB = BA = for  

general n .  The most direct way is probably to notice  

that when you multiply B  of this form on the left of  

any matrix, you are performing a sequence of row  

operations on that matrix:   

   * is unchanged, 

   *  

   *  

    ... 

    *  

Whe you perform these row operations on the matrix  A,  

for each  i > 1,  the  i th  row of  A  contains 1's in columns 

1, 2, ... i,  while the (i - 1)st  row contains 1's in columns  

1,2, ..., i-1.  The row operation   

the first i - 1  1's and leaves the  1  in row  i  and column i . 

Hence the product  BA  is the matrix   

It actually follows that  AB = too by the Invertible 

Matrix Theorem.  However, we can also prove this as follows. 

Similarly to what we said above about multiplying by B,   

when you multiply  A  on the left of any matrix,  

you are also doing a sequence of row operations: 

    * is unchanged 

    *  

    * (two operations) 

    ... 

     * (i-1 operations in all) 

    ...  

     * n-1  operations in all) 

When you apply these operations to the matrix B,  the pairs 

1, -1 in each column before the last cancel out and you are left  

with the identity matrix. 

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