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MATH 340; EXAM # 1, February 28, 2005 (R.A.Brualdi)
Discussion Section (circle one): Mon 8:50 Mon 12:05 Wed 8:50 Wed 12:
Name: These R. Solutions
- (12 points) Let A be the matrix
.
Let x be a 4 × 1 matrix (a column vector) (x 1 , x 2 , x 3 , x 4 )T^ , let y be a 1 × 3 matrix (y 1 , y 2 , y 3 ).
(i) Express Ax as a linear combination of the columns of A:
x 1
+ x 2
+ x 3
+ x 4
(ii) Express yA as a linear combination of the rows of A:
y 1
[ 1 2 3 4
]
[ 2 4 1 3
]
[ 3 5 1 2
]
(iii) Is yAx in the row space, column space, both the row and column space or neither? Circle one. NEITHER: yAx is a scalar (a 1 × 1 matrix).
- (14 points) Let A be a 4 × 3 matrix.
(i) Let E be the elementary matrix such that EA interchanges rows 2 and 4 of A. What is E and, if E is invertible, what is the inverse of E?
E =
We have E−^1 = E. (ii) Now let E be the elementary matrix such that EA is obtained from A by adding 10 times row 2 of A to row 4. What is E and, if E is invertible, what is the inverse of E?
E =
E−^1 =
- (15 points) Let A be an n × n invertible matrix.
(i) Define what it means for A to be invertible. That there exists a matrix B such that AB = BA = In. (ii) Prove that A^2 is invertible if A is invertible. Since A is invertible we know A−^1 exists. Consider the matrix B = (A−^1 )^2. Then using properties of matrix multiplication, one easily checks (do it!) that AB = BA = In One could also use determinants as in part (iii). (iii) If A^2 is invertible, is A invertible? Why or why not? We know that A is invertible if and only if det A 6 = 0. We also know that det A^2 = (det A)^2 since the deteminant is multiplicative. So since det A^2 6 = 0, det A 6 = 0 and A is invertible.
- (18 points) Let A be a 5 × 6 matrix whose reduced row echelon form is
(i) What is the dimension of the row space of A. What is a basis for it. This dimension is 3 and the first three rows of the matrix above form a basis. (ii) What is the dimension of the column space of A. What is a basis for it? This dimension is also 3 and columns 1, 3 and 5 of A form a basis. (iii) What is the dimension of the null space of A. What is a basis for it? This dimension is 6 = 3 = 3 and a basis can be found as follows. the above rref tells us that the solutions of Ax = 0 (the null space of A) is:
x 1 = − 2 c − 3 d − 2 f x 2 = c x 3 = − 5 d − 4 f x 4 = d x 5 = − 6 f x 6 = f
where d, e, f are arbitrary. Separating c, d and e, we get as a basis:
(− 2 , 1 , 0 , 0 , 0 , 0)T^ , (− 3 , 0 , − 5 , 1 , 0 , 0)T^ , (− 2 , 0 , − 4 , 0 , − 6 , 1)T^.
Hence the (1, 3)-entry of B−^1 equals (^) −− 157 = 157. Do the columns of B form a basis of R^3? Why or why not? Yes, because det B 6 = 0 and hence the columns of B are linearly independent, Since R^3 has dimension 3, these columns must form a basis of R^3.
