University Princeton Department of Mathematics
MAT 202 Linear Algebra with Applications QUIZ 1 Spring 2006 SOLUTIONS
(1) Solve the following system: x1 + 2x2 + x3 + x4 = 1 2x1 + 2x2 - x3 - x4 = 0 x1 - 2x2 - 5x3 + x4 = 0 What is the rank of the corresponding coefficient matrix and what is the rank of the corresponding augmented matrix?
Solution: We apply Gauss-Jordan elimination to the system written in matrix notation: 2 3 2 3 1 2 1 1 1 1 2 1 1 1 4 2 2 -1 -1 0 5 4 0 -2 -3 -3 -2 5 1 -2 -5 1 0 0 -4 -6 0 -1 3 3 2 2 1 0 -2 0 0 1 0 -2 -2 -1 1 3 3 3 1 54 0 1 0 4 5 4 0 1 2 2 2 0 0 0 6 3 0 0 0 1 1 2 The last matrix shows that the solutions of the system are the (x1 , x2 , x3 , x4 ) satisfying 8 < x1 - 2x3 = 0 1 x2 + 3 x3 = 4 2 : 1 x4 = 2 Taking x3 as a free variable, the 2 3 2 x1 2 6 x2 7 6 - 3 6 7=6 2 4 x3 5 4 1 x4 0 solutions form the vectors 3 2 3 0 1 7 7 6 7t + 6 4 7 , tR. 5 4 0 5
1 2
The rank of a matrix is the number of leading ones in its reduced rowechelon form (rref). The leading ones are boldfaced above. As there are three leading ones in the rref of the coefficient matrix and in the rref of the augmented matrix, we see that both of these matrices have rank 3.
(2) For what values of k does the following system: x1 - x2 = 1 x1 + kx2 - 4x3 = 3 -x1 + 2x2 - (1 + k)x3 = 0 have (a) infinitely many solutions? (b) no solutions? (c) exactly one solution? Justify your answers and provide a brief geometric interpretation.
2
Solution: We use Gauss-Jordan elimination: 3 3 2 2 1 1 1 -1 0 1 -1 0 2 5 3 54 0 k+1 4 1 -4 k -4 0 1 -(1 + k) 1 -1 2 -(1 + k) 0 3 3 2 2 1 1 1 -1 0 1 -1 0 1 5 1 -(1 + k) 1 5 4 0 1 -(1 + k) 4 0 2 0 k+1 -4 0 0 -4 + (1 + k)2 1 - k (a) In order to have infinitely many solutions we need the last row to be zero: -4 + (1 + k)2 = 0 and 1 - k = 0, that is k = 1. In this case the three planes intersect along one line. (b) To have no solutions we need to be short of leading ones, i.e., -4 + (1 + k)2 = 0, yet the last righthand side to be nonzero, i.e., 1 - k = 0, for the last equation to be inconsistent, that is k = -3. In this case the three planes have empty intersection. (c) To have exactly one solution we need three leading ones, therefore we must have -4 + (1 + k)2 = 0, that is k = 1 and k = -3. In this case the three planes intersect in exactly one point.
(3) Find the matrix A of the orthogonal projection onto the plane x1 - x2 + x3 = 0. Describe geometrically the transformation given A2 by . Is A invertible?
- Solution: The vector = (1,-1,1) has unit length and is orthogonal to n 3 the plane x1 - x2 + x3 = (1, -1, 1) (x1 , x2 , x3 ) = 0. The orthogonal projection onto this plane is the transformation proj (- ) = - - (- - )- . x x x n n
V
In matrix notation this is 3 3 2 2 2 3 2 2 x1 x1 1 3 1 A 4 x2 5 = 4 x2 5- (x1 -x2 +x3 ) 4 -1 5 = 4 1 3 3 x3 x3 1 -1 3 |
1 3 2 3 1 3
-1 3
1 3 2 3
{z
A
3 x1 5 4 x2 5 x3 } 32
Being a projection, the square of A represents the same projection. In general, the composition of a projection with itself is the same projection: projV (projV (- )) = projV (- ) - (projV (- ) - )- x x x n n - - (- - )- - ((- - (- - )- ) - )- = x x n n x x n n n n | {z } 0 - - (- - )- = x x n n = proj (- ) . x
V
The matrix A is not invertible as all the vectors proportional to - (vectors n orthogonal to the plane) project to zero, so its kernel is more than {0}.
(4) Find the inverse of the matrix 0 0 A= 4 1 0 0 1 0 0 1 2 0 1 0 3 0
3
Solution: Changing the order of rows in the first step and then using the leading ones in the first, third and fourth rows to also bring the second row to the reduced row-echelon form, we obtain 3 3 2 2 1 1 1 1 7 7 6 6 1 1 1 7 76 4 1 2 3 6 5 5 4 4 4 1 2 3 1 1 1 1 1 1 1 2 1 1 6 -3 -2 1 -4 1 6 4 1 1 1 1 where the omitted entries are zero. We conclude that 2 3 0 0 0 1 6-3 -2 1 -47 7 A-1 = 6 40 1 0 05 1 0 0 0
3 7 7 5
(5) Determine whether the following statements are true or false. No explanations, no credit. (a) Given two square matrices A and B, if AB is invertible, then so is A.
Solution: TRUE Let C be the inverse of AB, i.e., (AB)C = C(AB) = I. By associativity A(BC) = I, which implies that A is invertible with inverse BC.
(b) The composition of a shear, a reflection and a rotation in the plane is invertible.
Solution: TRUE A shear S, a reflection R and a rotation T are invertible transformations, hence their composition (in any order) is invertible. In particular, (SRT )-1 = T -1 R-1 S -1 .
(c) There exist matrices A of size 3 2 and B of size 2 3 such that AB = I3 is the identity matrix.
Solution: FALSE Since B is of size 2 3, we can find a nonzero vector - R3 such v - = - (the corresponding system has 2 equations for 3 variables, that B v 0 hence cannot have a unique solution, and zero is always a solution for a - homogeneous system). Therefore (AB)- = A(B - ) = 0 which shows v v that AB cannot be the identity matrix.