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MATH1025.03M


FW00


TERM TEST 2


Answers

1. (10 marks) Find the inverse of the matrixby using Gauss-Jordan elimination.

Indicate the specific elementary row operation used at each step.

->R3 -> R3 - 3 R1

->R2 -> (1/2) R2; R3 -> R3 - 4 R2

->R1 -> R1 + R3; R2 -> R2 - 3 R3

Thus the inverse matrix is


2. (6 marks) Given a square matrix A:

(a) Show that (AAT) is symmetric;

(AAT)T = (AT)TAT = AAT Hence AAT is symmetric.


(b) If A is invertible and AB = AC show that B = C.

If A is invertible, A-1 exists. Thus A-1(AB) =A-1(AC) or (A-1A)B = (A-1A)C which gives IB = IC or B = C


3. (10 marks) Given u = (2, -1, 5), v = (4, 3, 0) and w = (3, 1, -2) find

(a) 2u + v - w =


(b) u.w = 6 - 1 -10 = -5


(c) vxw == -6i + 8j - 5k or (-6, 8, -5)


(d) ||u|| =


(e) projuw = (u.w)/||u||2 u = -5/30 u = (-1/3, 1/6, -5/6) using the results from pats (b) and (d).


(f) the area of the parallelogram formed by v and w = || vxw || =using the results from part (c).

4. (6 marks) Use gaussian elimination to evaluate the determinant of.

Indicate the specific elementary row operation used at each step.


= 2x13 = 26

R1 -> (1/2) R1 and multiply the determinant by 2

R2 -> R2 - 3 R1; R3 -> R3 - 4 R2

Note in evaluating the last determinant we used the product of the diagonal elements.


5. (7 marks) Find the eigenvalues and eigenvectors of the matrix.

The characteristic polynomial is (s-2)(s-4) - 24 = s2 - 6s - 16 with roots 8 and -2.

Eigenvector for 8 is solution of

(2-8)x + 6y = 0

4x + (4-8)y = 0

which has solution x = y. Thus we have the eigenvector (1, 1) or any multiple of this.

Eigenvector for -2 is solution of

(2+2)x + 6y = 0

4x + (4+2)y = 0

which has solution x = -3/2 y. Thus we have the eigenvector (3, -2) or any multiple of this.


6. (4 marks) If A is an nxn invertible matrix with det(A) = 3 then evaluate

(a) det(AT) = 3; AT and A have the same determinant

(b) det(2A) = 2nx3

(c) det(4A-1) = 4nx1/3; det(A-1) = 1/det(A)

(d) det(AA-1) = det(I) = 1


7. (3 marks) Find the parametric equation of the line through the point P = (2, -2, 1) parallel to the vector v = (-1, 1, 4).

The line is given by x = OP + tv. In component form this is

x = 2 + (-1)t

y = -2 + t

x = 1 +4t


8. (2 marks) Find the plane through the point P = (5, -3, 1) with normal vector n = (-2, 3, 2).

The plane is given by n.x + d = 0, i.e. -2x + 3y + 2z + d = 0. We find d by substituting the coordinates of the point P into the equation giving d = 17 so the equation is -2x + 3y + 2z + 17= 0


9. (2 marks) Find the distance between the point (2, -4, 1) and the plane

4x + 6y - 2z + 7 = 0.

To find the distance we substitute the coordinates of the point into the equation of the plane, divide by the length of the normal vector to the plane and take the absolute value of the result. Here the normal vextor is (4, 6, -2) so the distance is



10. (4 marks) Indicate whether each of the following statements is true or false.

If A is an invertible matrix then

(i) the system of linear equations Ax = b has a unique solution: true

(ii) the homogeneous system Ax = 0 has a non-trivial solution: false

(iii) det(A) = 0 : false

(iv) the reduced row echelon form of A is the unit matrix I : true