MATH 244 -- Linear Algebra 

Problem Set 1 Solutions 

 

Section 1.1: 

 

1.1/30.  Multply row 2 by -1/2.  The opposite operation is to multiply the 

same row 2 by -2. 

 

1.1/32.  (that is, replace row 3 by row 3 plus 3 times row 2). 

The opposite operation is to replace row 3 by row 3 minus 3 times row 2. 

 

1.1/33.  The system of equations is 

 

            4*T[1]-T[2]-T[4] = 30 

            -T[1]+4*T[2]-T[3] = 60 

            -T[2]+4*T[3]-T[4] = 70 

            -T[1]-T[3]+4*T[4] = 40 

 

1.1/34  The solution is 

> with(linalg); -1
 

> A := matrix([[4, -1, 0, -1, 30], [-1, 4, -1, 0, 60], [0, -1, 4, -1, 70], [-1, 0, -1, 4, 40]]); 1
A := matrix([[4, -1, 0, -1, 30], [-1, 4, -1, 0, 60], [0, -1, 4, -1, 70], [-1, 0, -1, 4, 40]]); 1
 

table( [( 1, 1 ) = 4, ( 4, 4 ) = 4, ( 2, 2 ) = 4, ( 3, 3 ) = 4, ( 4, 3 ) = -1, ( 1, 2 ) = -1, ( 3, 2 ) = -1, ( 3, 1 ) = 0, ( 2, 3 ) = -1, ( 2, 5 ) = 60, ( 4, 5 ) = 40, ( 3, 4 ) = -1, ( 2, 1 ) = -1, ( ... 

> gaussjord(A)
 

table( [( 1, 1 ) = 1, ( 4, 4 ) = 1, ( 2, 2 ) = 1, ( 3, 3 ) = 1, ( 4, 3 ) = 0, ( 1, 2 ) = 0, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 0, ( 2, 5 ) = 55/2, ( 4, 5 ) = 45/2, ( 3, 4 ) = 0, ( 2, 1 ) = 0, ( 3,... 

The values of  T[1], T[2], T[3], T[4]are the entries in the last column. 

Section 1.2: 

 

1.2/4. 

> A := matrix([[1, 3, 5, 7], [3, 5, 7, 9], [5, 7, 9, 1]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 5, ( 3, 3 ) = 9, ( 1, 2 ) = 3, ( 3, 2 ) = 7, ( 3, 1 ) = 5, ( 2, 3 ) = 7, ( 3, 4 ) = 1, ( 2, 1 ) = 3, ( 1, 3 ) = 5, ( 1, 4 ) = 7, ( 2, 4 ) = 9 ] ) 

> gaussjord(A); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 1, ( 3, 3 ) = 0, ( 1, 2 ) = 0, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 2, ( 3, 4 ) = 1, ( 2, 1 ) = 0, ( 1, 3 ) = -1, ( 1, 4 ) = 0, ( 2, 4 ) = 0 ] ) 

The pivot columns are columns 1, 2, and 4. 

 

1.2/10 

 

> A := matrix([[1, -2, -1, 3], [3, -6, -2, 2]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = -6, ( 1, 2 ) = -2, ( 2, 3 ) = -2, ( 2, 1 ) = 3, ( 1, 3 ) = -1, ( 1, 4 ) = 3, ( 2, 4 ) = 2 ] ) 

> gaussjord(A); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 0, ( 1, 2 ) = -2, ( 2, 3 ) = 1, ( 2, 1 ) = 0, ( 1, 3 ) = 0, ( 1, 4 ) = -4, ( 2, 4 ) = -7 ] ) 

The general solution is  

where s  is any real number. 

 

 

1.2/12 

> A := matrix([[1, -7, 0, 6, 5], [0, 0, 1, -2, -3], [-1, 7, -4, 2, 7]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 0, ( 3, 3 ) = -4, ( 1, 2 ) = -7, ( 3, 2 ) = 7, ( 3, 1 ) = -1, ( 2, 3 ) = 1, ( 2, 5 ) = -3, ( 3, 4 ) = 2, ( 2, 1 ) = 0, ( 3, 5 ) = 7, ( 1, 3 ) = 0, ( 1, 4 ) = 6, ( 2, 4... 

> gaussjord(A); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 0, ( 3, 3 ) = 0, ( 1, 2 ) = -7, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 1, ( 2, 5 ) = -3, ( 3, 4 ) = 0, ( 2, 1 ) = 0, ( 3, 5 ) = 0, ( 1, 3 ) = 0, ( 1, 4 ) = 6, ( 2, 4 )... 

The general solution is 

where s, t are any real numbers. 

 

1.2/16 a.  System is consistent and solution is unique. 

          b.  System is consistent, but solution is not unique (x[2]is a free variable). 

 

1.2/18  Replace row 2 by row 2, plus (-5) times row 1.  The result is 

This augmented matrix represents an inconsistent 

system if  h = -15.  It is consistent for all other values of  h, since 

in all other cases, we could divide through on the second row  

by  h + 15 and solve for x[1] and x[2]. 

 

1.2/20.  In matrix form, the system is 

 

 Replace row 2 by row 2 plus (-3) times row 1: 

 

 

a)  The system has no solution (is inconsistent) if h = 9, k <> 6 

b)  The system has a unique solution if can be anything). 

c)  The system has infinitely many solutions if h = 9and  

    k = 6. 

 

1.2/23.  Yes.  The coefficient matrix is the part of the augmented 

matrix corresponding to the left-hand sides of the equations. 

If there are three pivot columns, we can solve for three of 

the variables in terms of the other two (free) variables.   

 

1.2/24.  No.  If the last column in the augmented matrix is 

a pivot column, then (in the rref) the row containing that  

pivot entry must look like:   0 0 0 0 1.  This gives an equation  

0 = 1, so the system is inconsistent. 

 

1.2/28.  It must be true that every column except the last is 

a pivot column (no free variables to get solution unique,  

and no pivot in the last column, so system is consistent). 

 

1.2/29.  If such a system is consistent, then there can  

be at most as many basic variables as equations in the 

original system.  This means there must be at least one 

free variable.  So there are infinitely many solutions. 

 

1.2/30.   The system with augmented matrix 

 

Matrix(%id = 138571604) is such an example.  The rref 

is 

so the system is inconsistent. 

 

1.2/31.  Yes an overdetermined system can be consistent. 

The system with augmented matrix 

 

Matrix(%id = 138171288)is an example.  The rref is  

(Note: the third row of the original matrix is the sum of the first  

two rows, so it cancels out completely with the row operations.) 

 

Section 1.3 

 

1.3/12.   

 

> A := matrix([[1, 0, 2, -5], [-2, 5, 0, 11], [2, 5, 8, -7]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 5, ( 3, 3 ) = 8, ( 1, 2 ) = 0, ( 3, 2 ) = 5, ( 3, 1 ) = 2, ( 2, 3 ) = 0, ( 3, 4 ) = -7, ( 2, 1 ) = -2, ( 1, 3 ) = 2, ( 1, 4 ) = -5, ( 2, 4 ) = 11 ] ) 

> gaussjord(A); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 1, ( 3, 3 ) = 0, ( 1, 2 ) = 0, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 4/5, ( 3, 4 ) = 1, ( 2, 1 ) = 0, ( 1, 3 ) = 2, ( 1, 4 ) = 0, ( 2, 4 ) = 0 ] ) 

There is a pivot in the last column, so this system has no solutions. 

The vector b is not in the linear combination of the a's. 

 

1.3/14. Same idea as 12: 

 

> A := matrix([[1, -2, -6, 11], [0, 3, 7, -5], [1, -2, 5, 9]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 3, ( 3, 3 ) = 5, ( 1, 2 ) = -2, ( 3, 2 ) = -2, ( 3, 1 ) = 1, ( 2, 3 ) = 7, ( 3, 4 ) = 9, ( 2, 1 ) = 0, ( 1, 3 ) = -6, ( 1, 4 ) = 11, ( 2, 4 ) = -5 ] ) 

> gaussjord(A); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 1, ( 3, 3 ) = 1, ( 1, 2 ) = 0, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 0, ( 3, 4 ) = -2/11, ( 2, 1 ) = 0, ( 1, 3 ) = 0, ( 1, 4 ) = 245/33, ( 2, 4 ) = -41/33 ] ) 

This system has a unique solution, giving the scalars 

in a linear combination of the columns of A that equals 

the vector b.   

 

1.3/16 and 20.  You can take any scalars and compute 

c[1]*v[1]+c[2]*v[2.] 

The vectors do not lie along the same line,  

so the Span is a plane in  

 

1.3/18. 

By looking at the 2nd and 3rd components of a 

linear combination c[1]*v[1]+c[2]*v[2] = y 

you can see  

so Then, from the first components, 

(You can also do this 

by reducing to echelon form and seeing when the  

system  is consistent, of course.) 

 

1.3/22.  We just need to force the b to not be 

a linear combination of the columns in A.  One 

easy way to do this is to give A a whole row of 

zeroes (say in row 3), but put a nonzero entry in  

the third row of b: 

 

 

(Note the problem just says "A has nonzero entries," 

which is true here, not "A has only nonzero entries." 

If you insist that A have only nonzero entries, then 

you can get an example, for instance, by making  

all the columns of A the same, but b not a scalar 

multiple of that vector: 

 

A = `.`(Matrix(%id = 136790748), b) and `.`(Matrix(%id = 136790748), b) = Vector[column](%id = 138148356) 

 

1.3/25 

a.  No.  This notation {a[1], a[2], a[3]}means the set consisting 

of the three columns of A (just those three vectors).  b 

is not one of those columns. 

 

b.  Yes: 

> A := matrix([[1, 0, -4, 4], [0, 3, -2, 1], [-2, 6, 3, -4]]); 1
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 3, ( 3, 3 ) = 3, ( 1, 2 ) = 0, ( 3, 2 ) = 6, ( 3, 1 ) = -2, ( 2, 3 ) = -2, ( 3, 4 ) = -4, ( 2, 1 ) = 0, ( 1, 3 ) = -4, ( 1, 4 ) = 4, ( 2, 4 ) = 1 ] ) 

> gaussjord(A)
 

table( [( 1, 1 ) = 1, ( 2, 2 ) = 1, ( 3, 3 ) = 1, ( 1, 2 ) = 0, ( 3, 2 ) = 0, ( 3, 1 ) = 0, ( 2, 3 ) = 0, ( 3, 4 ) = -2, ( 2, 1 ) = 0, ( 1, 3 ) = 0, ( 1, 4 ) = -4, ( 2, 4 ) = -1 ] ) 

There is a solution of this system (unique in fact).  So 

b is in W.  There are infinitely many vectors in W because 

we can use any real scalars in the linear combinations. 

 

c.  If then  

This shows a[1]is in W.