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This is the Exam Key of Applied Linear Algebra which includes Homogeneous Coordinates, Linear Equations, Precise Description, Linearly Independent etc. Key important points are: Determinant, Matrix, Real Number, Values, Invertible, Formula, Valid, Basis For Nul, Basis For Col, Rank
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Math 232 Second Midterm Solution March 5, 2007
Answer
We compute the determinant by expansion with respect to the first row:
det
(^) = 2 det
1 c 0 1 1 c
Answer
Again we expand by first row to get
det(A) = det
1 c 1 1
= 1 − c
(b) (3 points) For which values of c is A invertible? Give a formula for A−^1 that is valid for those values of c.
Answer
A square matrix is invertible if and only if its determinant is non-zero, so for A this is the case whenever c 6 = 1. In those cases, we can use the adjugate formula to compute the inverse of A:
A−^1 =
det(A)
c 11 c 21 c 31 c 12 c 22 c 32 c 13 c 23 c 33
where cij = (−1)i+j^ det Aij and Aij is A with the i-th row and the j-th column removed. We thus get
c 11 = det
c 0 1 c
= c^2 c 12 = − det
1 c
= −c c 13 = det
1 c 1 1
= 1 − c
c 21 = − det
1 c
= 1 c 22 = det
1 c
= − 1 c 23 = − det
c 31 = det
c 0
= −c c 32 = − det
= 1 c 33 = det
1 c
and hence
A−^1 =
1 − c
c^2 1 −c −c − 1 1 1 − c 0 0
c^2 1 −c
1 1 −c
−c 1 −c −c 1 −c
− 1 1 −c
1 1 −c 1 0 0
Answer
(i) The zero vector (i.e., zero polynomial) should be in H (ii) If f 1 (t), f 2 (t) ∈ H then f 1 (t) + f 2 (t) ∈ H. (iii) If f 1 (t) ∈ H and c ∈ R then cf 1 (t) ∈ H.
(b) (3 points) Prove that H is a subspace of P 3.
Answer
(i) If f (t) = 0, the zero polynomial, then f (t) = 0 = f (−t), so the zero polynomial is in H. (ii) Suppose f 1 (t), f 2 (t) ∈ H and let h(t) = f 1 (t) + f 2 (t). Then
h(−t) = f 1 (−t) + f 2 (−t) = f 1 (t) + f 2 (t) = h(t)
so H is closed under vector addition. (iii) If f 1 (t) ∈ H and c ∈ R then cf 1 (t) ∈ H. Suppose f 1 (t) ∈ H, c ∈ R and let h(t) = cf (t). Then h(−t) = cf (−t) = cf (t) = h(t), so H is closed under scalar multiplication as well.
(c) (3 points) Give a basis of H. Explain why your answer is correct.
Answer
In general, an element of P 3 can be written as f (t) = a 0 +a 1 t+a 2 t^2 +a 3 t^3 for a 0 , a 1 , a 2 , a 3 ∈ R. The f (t) = f (−t) implies that
a 0 + a 1 t + a 2 t^2 + a 3 t^3 = a 0 − a 1 t + a 2 t^2 − a 3 t^3 ,
so a 0 = a 0 , a 1 = −a 1 , a 2 = a 2 , a 3 = −a 3. In other words, a 1 = a 3 = 0. That means that any element of H can be written as a linear combination of 1 and t^2. Since these are linearly independent, we see that { 1 , t^2 } is a basis for H.
H =
x 1 x 2 x 3
(^) : R^3 : x 1 + x 2 = x 3
(a) (3 points) Show that B =
is a basis for H.
Answer
First we show that B is linearly independent: If x 1
(^) + x 2
(^) = 0 then x 1 = 0 from
the first coordinate and x 2 = 0 from the second coordinate. Second, we check that B ⊂ H: Clearly, the vectors in B satisfy the equation x 1 + x 2 = x 3. Together with the first bit, this establishes that dim H ≥ 2 and that if dim H = 2 then B must span H and hence is a basis.
Third, note that H 6 = R^3 , because the vector
(^) ∈/ H. Hence, dim H < dim R^3 = 3, so
indeed dim H = 2.
(b) (2 points) Describe in words what the coordinate vector [v]B of a vector v ∈ H is.
Answer
The coordinate vector [v]B is the unique vector
c 1 c 2
such that v = c 1 b 1 + c 2 b 2 , where b 1 , b 2 are the elements of B.
