Page 353 - 35Linear Algebra

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Now solve UX = W by back substitution

x + y + z + w = 1, y + z + w = 0, z + w = 0

⇒ w = µ (arbitrary), z = −µ, y = 0, x = 1 .

   
 x 1

 
y 0
 
   
The solution set is   =   : µ ∈ R
 z  −µ  
 
 
y µ
 
5. First

1 2
det = −2 .
3 4
All the other determinants vanish because the first three rows of each matrix
are not independent. Indeed, 2R 2 − R 1 = R 3 in each case, so we can make
row operations to get a row of zeros and thus a zero determinant.
 
x
3
6. If U spans R , then we must be able to express any vector X =   ∈ R 3
y
z
as
1
         
1 1 a 1 1 a c
1
c
X = c 1   + c 2  2  + c 3   =  0 2 1   2  ,
0
1 −3 0 1 −3 0 c 3
1
3
2
for some coefficients c , c and c . This is a linear system. We could solve
3
1
2
for c , c and c using an augmented matrix and row operations. However,
3
3
since we know that dim R = 3, if U spans R , it will also be a basis. Then
1
2
3
the solution for c , c and c would be unique. Hence, the 3×3 matrix above
must be invertible, so we examine its determinant
 
1 1 a
det  0 2 1  = 1.(2.0 − 1.(−3)) + 1.(1.1 − a.2) = 4 − 2a .
1 −3 0
3
Thus U spans R whenever a 6= 2. When a = 2 we can write the third vector
in U in terms of the preceding ones as
     
2 1 1
3 1
1
0
  =   +  2  .
2 2
0 1 −3
(You can obtain this result, or an equivalent one by studying the above linear
system with X = 0, i.e., the associated homogeneous system.) The two
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