IB HL Math Summer Homework (2006): Part IV (Systems of Equations) Solutions
1)
 2  1  9  x   7 

   
3  y    1 
1 2
 2 1  3  z   k 

   
 2  1  9  x   7 

  

 0 2.5 7.5  y     2.5 
0 2
6  z   k  7 

7
 2  1  9  x  


  

 0 2.5 7.5  y     2.5 
0 0
0  z   k  7  2 

If the system has any solutions at all, it must be the case that k-7+2 is equal to 0. Hence,
the value of k for which the system has an infinite number of solutions is k=5.
2)
 4  1 2  x   1 

   
 2 3 0  y     6 
 1  2 a  z   3.5 

   
1
2  x   1 
4

  

 1  y     6.5 
 0 3.5
 0  1.75 a  .5  z   3.25 

  

2  x   1 
 4 1

  

 0 3.5  1  y     6.5 
 0 0 a  1 z   0 

  

We see that if a-1 is NOT 0, then z has a unique solution which will end up forcing both
x and y to have unique solutions as well. Thus, for this system to NOT have a unique
solution, a must equal 1.
3)
1

2
1

1

0
0

1

0
0

1  x   k 
   
1 4  y    6 
 4 5  z   9 
2
1  x   k 
  

 3 2  y    6  2k 
 6 4  z   9  k 
2
1  x  
k

  

 3 2  y   
6  2k





0 0  z   9  k  2(6  2k ) 
2
Since the last equation here drops out, we find that the solution is NOT unique. We can
solve for the value of k for which the equations are consistent:
9 – k – 2(6 – 2k) = 0
9 – k – 12 + 4k = 0
3k = 3
k = 1.
For this value of k, we find that the second adjusted equation of the system is:
-3y + 2z = 4
Solving for z we find: z 
3y  4
.
2
Now, plug this into the first original equation and once again solve for z in terms of x:
x + 2y + z = 1
2z  4
x  2(
) z 1
3
3x  2(2 z  4)  3z  3
3x  4z  8  3z  3
7 z  3x  11
 3 x  11
 3x  11 3 y  4 z  0
z


, so the final solution is
7
7
2
1
4)
 1 4 1


 1 3

det x 2 2   det
 x 3
 x2 2 1


(1)(2)(1)  (4)(2)( x 2 )  (1)( x)(2)  ( x 2 )( 2)(1)  (2)(2)(1)  (1)( x)(4)  (1)(3)  ( x)(3)
2  8 x 2  2 x  2 x 2  4  4 x  3  3x
6x 2  x  5  0
(6 x  5)( x  1)  0
x  5 or x  1
6
5)
 1  1 2  x    1

   
 4 1 1  y    13 
 5  1 8  z   5 

   
 1  1 2  x    1

   
 0 5  7  y    17 
 0 4  2  z   10 

   
 1  1 2  x    1 

  

 0 5  7  y    17 
 0 0 3.6  z    3.6 

  

 1  1 2  x    1

   
 0 5  7  y    17 
0 0
1  z    1

 1  1 2  x    1

   
 0 5 0  y    10 
 0 0 1  z    1

   
 1 0 0  x   3 

   
 0 1 0  y    2 
 0 0 1  z    1

   
6) Plug in the three points to yield the following three equations:
f(-1) = – a + b – c + d = 0
f(0) =
d=2
f(1) = a + b + c + d = 0
First, substitute for d into the first and third equations:
f(-1) = – a + b – c + 2 = 0
f(1) = a + b + c + 2 = 0
Now, add these equations together:
2b + 4 = 0, so b = -2.
7) Adding all of the five equations together yields the following:
6 x1  6 x2  6 x3  6 x4  6 x5  186
x1  x2  x3  x4  x5  31
Now, solve for x4 by subtracting this equation from the four original equation to yield:
x4  17
Similarly, solve for x5 by subtracting this equation from the fifth original one:
x5  65 .
Thus, 3x4 + 2x5 = 3(17) + 2(65) = 51 + 130 = 181.
8) We notice that there aren't seven equations, thus, the system doesn't have a unique
solution, but nonetheless, there is a unique value that the expression
16 x1  25x2  36 x3  49 x4  64 x5  81x6  100 x7 must evaluate to.
Somehow, we must find some linear combination of the given equations to equal the
given expression on one side of the equal sign. Let us multiply the first equation by a
constant a, the second by a constant b and the third by a constant c.
Based on the pattern, we must have the following:
a(n-1)2 + b(n)2 + c(n+1)2 = (n+2)2.
Now, multiply out and equate coefficients of both sides:
an2 -2an + a + bn2 +cn2 +2cn + c = n2 + 4n + 4
(a+b+c)n2 +(-2a+2c)n + (a+c) = n2 + 4n + 4
Thus, we arrive at the following system of equations by equating coefficients of both
quadratics in the equation above:
a+b+c = 1
-2a + 2c = 4
a + c = 4, so 2a + 2c = 8
Adding the last two equations, we find that 4c = 12 and c = 3. Plugging back into the first
equation we find that a = 1. Finally, we find that b = -3 by plugging in both of these
values into the first equation.
This is the derivation to show the following:
( x1  4 x2  9 x3  16 x4  25x5  36 x6  49 x7 )  1
 3(4 x1  9 x2  16 x3  25x4  36 x5  49 x6  64 x7 )  36
3(9 x1  16 x2  25x3  36 x4  49 x5  64 x6  81x7 )  369
--------------------------------------------------------------------16 x1  25x2  36 x3  49 x4  64 x5  81x6  100 x7  369  36  1  334
So basically, if we multiply the first equation by 1, the second by -3 and the last one by 3,
and add the three equations, the left-hand side of the resulting equation is the expression
for which we need to solve. The right-hand side of this equation is just 334, the desired
answer.
9) Let A = log10x, B = log10y, and C= log10z. Substitute into the given equations:
log 10 2000  log 10 x  log 10 y  (log 10 x)(log 10 y)  4
log 10 2  log 10 y  log 10 z  (log 10 y)(log 10 z )  1
log 10 z  log 10 x  (log 10 z )(log 10 x)  0
log 10 2  log 10 1000  A  B  AB  4
log 10 2  B  C  BC  1
C  A  CA  0
log 10 2  3  A  B  AB  4
log 10 2  A  B  AB  1
At this point, quickly note that neither A nor B can be 1 because plugging these values
into the first two equations yields a false equation.
Solve for C in terms of A:
C  A  CA  0
A  CA  C
A  C ( A  1)
A
C
, which is valid since we know that A  1 .
A 1
Now, go back to the two other equations:
log 10 2  A  B  AB  1
log 10 2  B  C  BC  1
And subtract the top equation from the bottom one:
C  A  BC  AB  0 , then substitute for C,
A
A
 A  B(
)  AB  0 , multiply through by A-1
A 1
A 1
A  A( A  1)  AB  AB( A  1)  0
A(1  ( A  1))  AB( A  1  1)  0
A(2  A)  AB( A  2)  0
 A( A  2)  AB( A  2)  0
A( A  2)( B  1)  0
In was previously established that B can not equal one. Thus, it follows that either A = 0
or A = 2.
Let's solve for B in both cases:
Plug in A = 0 into log 10 2  A  B  AB  1 :
log 10 2  0  B  0  1
B  log 10 10  log 10 2
B  log 10 5 , the corresponding solution for y is y  5 .
Now, plug in A = 2 into log 10 2  A  B  AB  1 :
log 10 2  2  B  2B  1
log 10 2 1  B
log 10 2  log 10 10  B
B  log 10 20 , the corresponding solution for y is y  20 .
The desired sum is 25.