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MATH 244 -- Linear Algebra, section 1 

Problem Set 4 Solutions 

February 23, 2007 

Section 2.3 

6.  After replacing by  we have 

We see that rows 2 and 3 are  

now proportional -- if we replace by  

the last row will be all zeros.  Hence   A   is not invertible. 

8.  There is a pivot in each of the 4 columns in this 

upper-triangular  4 x 4 matrix, hence it is invertible. 

14.  A lower-triangular matrix is invertible if and only 

if all of its diagonal entries are nonzero.  This follows 

from statement c in the Invertible Matrix Theorem (Theorem 

8 in this section).  In order for there to be  n  pivots the 

diagonal entries must all be nonzero.  Conversely, if all 

the diagonal entries are nonzero, then there are  n  pivots. 

15.  No.  By the Invertible Matrix Theorem part e,  

if A  is invertible, then its columns are linearly independent. 

If there are two equal columns, the collection of columns 

is not linearly independent, so the matrix is not invertible. 

18.  No.  If  C x = v  is consistent for all  v,  then the mapping 

defined by  C   is onto (part i of the Invertible Matrix Theorem). 

As a result, part f also holds -- the mapping defined by  C 

is one-to-one, which says that C x = v  has a unique solution 

x   for all v.   

22.  If the system  Hx = c  is inconsistent for some  c, 

then part i of the Invertible Matrix Theorem is not true. 

The equivalence of the statements there means that  

all of the other parts fail for this  H.   In particular,  

the negation of part d  holds, and there is a nontrivial 

solution of the homogeneous system   

26.  If the columns of  A  are linearly independent, then 

part e of the Invertible Matrix Theorem holds, so A 

is an invertible matrix (exists).  Since    exists, 

is also an invertible matrix:    

But then the Invertible Matrix Theorem implies that statement 

h holds for  

30.  All of the statements in the Invertible Matrix Theorem 

hold for  A  and the linear mapping defined by  A. 

36.   Let  A  be the standard matrix of T.   Since T 

is onto, part i of the Invertible Matrix Theorem holds,  

hence all the other parts hold too.  In particular,  A 

is an invertible matrix.  Since  exists, we can  

use it to define a linear transformation too, and this 

is the inverse mapping  is also invertible 

(its inverse is A).  So all the parts of the Invertible 

Matrix Theorem also apply to and   

In particular,  is both onto and one-to-one. 

(Of course, this also follows from general facts 

about when inverse mappings exist and what 

there properties are.) 

38.  The answer is No.  The reason is that the given 

information that for some  says 

that T  is not one-to-one.  Hence all of the statements 

in the Invertible Matrix Theorem are false for  T.   

In particular, T is not onto.  

Section 2.8 

8. 

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The fact that this system is consistent shows that p  

is in the column space of A.   

16.  Not a basis for since the set is not linearly  

independent.  Note that if  and are the given  

vectors, then  

18. 

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This shows that the given set is linearly independent.   

The Invertible Matrix Theorem says that the matrix 

with these columns is invertible, so the columns  

span as well.  Hence, they give a basis for  

26.  The pivot columns are columns 1,2,4.  Hence 

(by Theorem 13 in this section),  columns 1,2,4 

of A  are a basis for Col(A): 

To find a basis for  Nul(A), continue and put the  

augmented matrix for Ax = 0  in rref: 

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The parametrization of the set of solutions is determined by the  

free variables  

The two vectors here are a basis for Nul(A). 

32.  It is not all of by equivalence of parts d and h of the  

Invertible Matrix Theorem). 

34.  The solution of  Px = b  is unique for all  b  

by the equivalence of parts d, f, g  of the Invertible 

Matrix Theorem. 

Section 2.9.   

6.  This just means to find the scalars such  

that  

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We see and   

10.  This is similar to number 26 from Section 2.8.  A basis 

for the column space consists of the pivot columns of  

A -- columns 1,2,4.  So dim Col(A) = 3.  Then  

dim Nul(A) = 2 (the number of free variables).  Basis 

for Nul(A): 

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(from parametrization of set of solutions of homogeneous system). 

16.  No.  Col(A) cannot equal , since the columns are vectors with 

four components.  The column space is a three-dimensional 

subspace of The dimension of  Nul(A)  is the number of  

free variables in the system.  Since there are 3 pivot columns 

out of 7 columns all together, there must be 4 free variables,  

so  dim Nul(A) = 4. 

20.  dim Nul(A) = 3  means there are 3 free variables. 

In a 4 x 5 matrix, that leaves 2 pivot columns, and two 

basis vectors for the column space.  The rank 

is the dimension of the column space, so rank(A) = 2.  

22.  If dim Span(let  { 

be a basis for that subspace.  Each can be written as a linear 

combination of the basis vectors, so  

         ... 

Consider a linear combination of the that adds up to the 

zero vector: 

substitute the above expressions, and collect  

the terms involving each of the  

Since the are linearly independent, this says 

   0 =  

   0 =  

   0 =  

   0 =  

This is a homogeneous system of 4 linear equations in 5 variables 

(the  

variables than equations.  Hence there is a nontrivial solution, and 

the  are linearly dependent. 

24.  Rank = 1 means that the columns span a one-dimensional 

space.  Each column is a scalar multiple of one of them.  Anything like 

 gives an example. 

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