Questions about the assignment may be sent to the TA, Lauren D'Elia at rldelia@math.ncsu.edu.
Each problem is worth 25 points. There are several ways to solve each problem, only 1 is shown.
1. Let
[3 4 1]
A = [2 6 3]
[0 1 8]
Compute A^2-3A and A^3+10I, assuming I is a 3x3.
PART A SOLUTION: (15 points)
[3 4 1] [3 4 1]
A^2 = [2 6 3] * [2 6 3]
[0 1 8] [0 1 8]
[17 37 23]
= [18 47 44]
[2 14 67]
[9 12 3]
3*A = [6 18 9]
[0 3 24]
[8 25 20]
A^2 – 3A = [12 29 35]
[2 11 43]
PART B SOLUTION: (10 points)
[3 4 1] [17 37 23]
A^3 = A*A^2 = [2 6 3] * [18 47 44]
[0 1 8] [2 14 67]
[125 313 312]
= [148 398 511]
[34 159 580]
[10 0 0]
10*I = [0 10 0]
[0 0 10]
[135 313 312]
A^3 + 10I = [148 408 511]
[34 159 590]
2. Let
[1 4]
A = [2 5] and
[3 0]
[0 2 1 5]
B = [4 -3 2 -1]
[1 2 0 -2]
Compute each of the following if possible and if not possible explain why.
a.(A^T)A
b.BB^T
c.(AB)^T
d.(B^T)A
e.(B^T)(A^T)
PART A SOLUTION: (5 points)
[1 2 3]
A^T = [4 5 0]
[1 4]
A = [2 5]
[3 0]
[14 14]
(A^T)A = [14 41]
PART B SOLUTION: (5 points)
[0 2 1 5]
B = [4 -3 2 -1]
[1 2 0 -2]
[0 4 1]
B^T = [2 -3 2]
[1 2 0]
[5 -1 -2]
[30 -9 -6]
BB^T = [-9 30 0]
[-6 0 9]
PART C SOLUTION: (5 points)
Not possible: A is a 3x2 and B is a 3x4. The number of columns in A is not equivalent to the number of rows in B.
PART D SOLUTION: (5 points)
[0 4 1]
B^T = [2 -3 2]
[1 2 0]
[5 -1 -2]
[1 4]
A = [2 5]
[3 0]
(B^T)A = [11 20]
[2 -7]
[5 14]
[-3 15]
PART E SOLUTION: (5 points)
Not possible: B^T is a 4x3 and A^T is a 2x3. The number of columns in B^T is not equivalent to the number of rows in A^T.
3. Find A^-1 using the Gauss-Jordan elimination method discussed in class (do not use the matrix inverse command in
Maple) if
[1 4 3]
A = [-1 -2 0]
[2 2 3]
SOLUTION:
[1 4 3| 1 0 0]
[-1 -2 0| 0 1 0]
[2 2 3| 0 0 1]
[1 4 3| 1 0 0]
--> [0 2 3| 1 1 0]
[0 -6 -3| -2 0 1]
[1 4 3| 1 0 0]
--> [0 2 3| 1 1 0]
[0 0 6| 1 3 1]
[1 4 0| ½ -3/2 -1/2]
--> [0 2 0| ½ -1/2 -1/2]
[0 0 6| 1 3 1]
[1 0 0| -1/2 -1/2 ½]
--> [0 2 0| ½ -1/2 -1/2]
[0 0 6| 1 3 1]
[1 0 0| -1/2 -1/2 ½]
--> [0 1 0| ¼ -1/4 -1/4]
[0 0 1| 1/6 ½ 1/6]
[-1/2 -1/2 ½]
A^-1 = [¼ -1/4 -1/4]
[1/6 ½ 1/6]
4. Let
[0 2 3 0]
A = [0 4 5 0]
[0 1 0 3]
[2 0 1 3]
[x1] [8]
x = [x2] b = [3]
[x3] [6]
[x4] [2]
a.Write down the matrix M=t*I-A where t is a variable and I is in 4x4 form.
b.Compute det(M) using cofactor expansion. First express the determinant as a sum of 3x3 cofactors and then
compute the determinant.
c. By using Cramer's Rule, express each component of the solution to Ax=b as a quotient of 4x4 determinants.
d. Compute each of the determinants from part b to find the unique solution to Ax=b.
PART A SOLUTION:(5 points)
[1 0 0 0] [0 2 3 0]
M = t x [0 1 0 0] - [0 4 5 0]
[0 0 1 0] [0 1 0 3]
[0 0 0 1] [2 0 1 3]
[t -2 -3 0]
= [0 t - 4 -5 0]
[0 -1 t -3]
[-2 0 -1 t – 3]
PART B SOLUTION:(10 points)
|t-4 -5 0| |-2 -3 0|
det(M) = t* |-1 t -3| - (-2) |t-4 -5 0|
|0 -1 t-3| |-1 t -3|
= t (t-4) |t -3| - t (-1) |-5 0| + 2 (-2) |-5 0| - 2 (-3) |t-4 0|
|-1 t-3| |-1 t-3| |t -3| |-1 -3|
= (t^2-4t)[(t^2-3t)-3] + t(-5t+15) – 4(15) + 6(-3t+12)
=t^4-4t^3-3t^3+12t^2-3t^2+12t -5t^2 +15t -60-18t +72
=t^4-7t^3+4t^2+9t +12
PART C SOLUTION:(5 points)
A*x = b:
[0 2 3 0] [x1] [8]
[0 4 5 0] * [x2] = [3]
[0 1 0 3] [x3] [6]
[2 0 1 3] [x4] [2]
Cramer's Rule:
x1 = det [8 2 3 0]
[3 4 5 0]
[6 1 0 3]
[2 0 1 3]
det [0 2 3 0]
[0 4 5 0]
[0 1 0 3]
[2 0 1 3]
x2 = det [0 8 3 0]
[0 3 5 0]
[0 6 0 3]
[2 2 1 3]
det [0 2 3 0]
[0 4 5 0]
[0 1 0 3]
[2 0 1 3]
x3 = det [0 2 8 0]
[0 4 3 0]
[0 1 6 3]
[2 0 2 3]
det [0 2 3 0]
[0 4 5 0]
[0 1 0 3]
[2 0 1 3]
x4 = det [0 2 3 8]
[0 4 5 3]
[0 1 0 6]
[2 0 1 2]
[0 2 3 0]
[0 4 5 0]
[0 1 0 3]
[2 0 1 3]
PART D SOLUTION: (5 points)
with(linalg):Warning, the protected names norm and trace have been redefined and unprotected
> A:=matrix(4,4,[0,2,3,0,0,4,5,0,0,1,0,3,2,0,1,3]);
[0 2 3 0]
[ ]
[0 4 5 0]
A := [ ]
[0 1 0 3]
[ ]
[2 0 1 3]
> det(A);
12
> x1:=matrix(4,4,[8,2,3,0,3,4,5,0,6,1,0,3,2,0,1,3]);
[8 2 3 0]
[ ]
[3 4 5 0]
x1 := [ ]
[6 1 0 3]
[ ]
[2 0 1 3]
> det(x1);
-195
> det(x1)/det(A);
-65/4
> x2:=matrix(4,4,[0,8,3,0,0,3,5,0,0,6,0,3,2,2,1,3]);
[0 8 3 0]
[ ]
[0 3 5 0]
x2 := [ ]
[0 6 0 3]
[ ]
[2 2 1 3]
> det(x2);
-186
> det(x2)/det(A);
-31/2
> x3:=matrix(4,4,[0,2,8,0,0,4,3,0,0,1,6,3,2,0,2,3]);
[0 2 8 0]
[ ]
[0 4 3 0]
x3 := [ ]
[0 1 6 3]
[ ]
[2 0 2 3]
> det(x3);
156
> det(x3)/det(A);
13
> x4:=matrix(4,4,[0,2,3,8,0,4,5,3,0,1,0,6,2,0,1,2]);
[0 2 3 8]
[ ]
[0 4 5 3]
x4 := [ ]
[0 1 0 6]
[ ]
[2 0 1 2]
> det(x4);
86
> det(x4)/det(A);
43/6