Section 6.1

Determinants

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Problem 6.1.1

Use the definition to evaluate det⁡(A)\operatorname {det}\left(\mathbf{A}\right) for each of the following matrices.

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(a)

A=(3−21−540216)\mathbf{A} = \begin{pmatrix} 3 & -2 & 1 \\ -5 & 4 & 0 \\ 2 & 1 & 6 \end{pmatrix}.

(b)

A=(211621−221)\mathbf{A} = \begin{pmatrix} 2 & 1 & 1 \\ 6 & 2 & 1 \\ -2 & 2 & 1 \end{pmatrix}.

(c)

A=(00α0β0γ00)\mathbf{A} = \begin{pmatrix} 0 & 0 & \alpha \\ 0 & \beta & 0 \\ \gamma & 0 & 0 \end{pmatrix}.

(d)

A=(a11a12a13a21a22a23a31a32a33)\mathbf{A} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}.

Problem 6.1.2

What is the volume of the parallelepiped generated by the three vectors x→1=(3,0,−4,0)T\overrightarrow {x}_{1} = (3,0,-4,0)^{T}, x→2=(0,2,0,−2)T\overrightarrow {x}_{2} = (0,2,0,-2)^{T}, and x→3=(0,1,0,1)T\overrightarrow {x}_{3} = (0,1,0,1)^{T}?

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Problem 6.1.3

Using Gaussian elimination to reduce A\mathbf{A} to an upper-triangular matrix, evaluate det⁡(A)\operatorname {det}\left(\mathbf{A}\right) for each of the following matrices.

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(a)

A=(123241144)\mathbf{A} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 1 \\ 1 & 4 & 4 \end{pmatrix}.

(b)

A=(135−1423−24)\mathbf{A} = \begin{pmatrix} 1 & 3 & 5 \\ -1 & 4 & 2 \\ 3 & -2 & 4 \end{pmatrix}.

(c)

A=(12−344812−82321−3−11−4)\mathbf{A} = \begin{pmatrix} 1 & 2 & -3 & 4 \\ 4 & 8 & 12 & -8 \\ 2 & 3 & 2 & 1 \\ -3 & -1 & 1 & -4 \end{pmatrix}.

(d)

A=(00−231012−112102−30)\mathbf{A} = \begin{pmatrix} 0 & 0 & -2 & 3 \\ 1 & 0 & 1 & 2 \\ -1 & 1 & 2 & 1 \\ 0 & 2 & -3 & 0 \end{pmatrix}.

(e)

A=(2−1000−12−1000−12−1000−12−1000−11)\mathbf{A} = \begin{pmatrix} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 2 & -1 & 0 \\ 0 & 0 & -1 & 2 & -1 \\ 0 & 0 & 0 & -1 & 1 \end{pmatrix}.

(f)

A=(111⋯1121⋯1113⋯1⋮⋮⋮⋱⋮111⋯n)\mathbf{A} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 1 & \cdots & 1 \\ 1 & 1 & 3 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & n \end{pmatrix}.

Problem 6.1.4

Use determinants to compute the rank of A=(13−2012−1−1625−6)\mathbf{A} = \begin{pmatrix} 1 & 3 & -2 \\ 0 & 1 & 2 \\ -1 & -1 & 6 \\ 2 & 5 & -6 \end{pmatrix}.

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Problem 6.1.5

Use determinants to find the values of α\alpha for which the following system possesses a unique solution.

(1α001−1α01)(x1x2x3)=(−347). \begin{pmatrix} 1 & \alpha & 0 \\ 0 & 1 & -1 \\ \alpha & 0 & 1 \end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \\ 7 \end{pmatrix}.
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Problem 6.1.6

If A\mathbf{A} is nonsingular, explain why det⁡(A−1)=1/det⁡(A)\operatorname {det}\left(\mathbf{A}^{-1}\right) = 1/\operatorname {det}\left(\mathbf{A}\right).

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Problem 6.1.7

Explain why determinants are invariant under similarity transformations. That is, show det⁡(P−1AP)=det⁡(A)\operatorname {det}\left(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\right) = \operatorname {det}\left(\mathbf{A}\right) for all nonsingular P\mathbf{P}.

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Problem 6.1.8

Explain why det⁡(A∗)=det⁡(A)‾\operatorname {det}\left(\mathbf{A}^{*}\right) = \overline{\operatorname {det}\left(\mathbf{A}\right)}.

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Problem 6.1.9
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(a)

Explain why ∣det⁡(Q)∣=1\left|\operatorname {det}\left(\mathbf{Q}\right)\right| = 1 when Q\mathbf{Q} is unitary. In particular, det⁡(Q)=±1\operatorname {det}\left(\mathbf{Q}\right) = \pm 1 if Q\mathbf{Q} is an orthogonal matrix.

(b)

How are the singular values of A∈Cn×n\mathbf{A} \in \mathbb {C}^{n \times n} related to det⁡(A)\operatorname {det}\left(\mathbf{A}\right)?

Problem 6.1.10

Prove that if A\mathbf{A} is m×nm \times n, then det⁡(A∗A)≥0\operatorname {det}\left(\mathbf{A}^{*}\mathbf{A}\right) \geq 0, and explain why det⁡(A∗A)>0\operatorname {det}\left(\mathbf{A}^{*}\mathbf{A}\right) > 0 if and only if rk⁡(A)=n\operatorname {rk}\left(\mathbf{A}\right) = n.

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Problem 6.1.11

If A\mathbf{A} is n×nn \times n, explain why det⁡(αA)=αndet⁡(A)\operatorname {det}\left(\alpha \mathbf{A}\right) = \alpha^{n}\operatorname {det}\left(\mathbf{A}\right) for all scalars α\alpha.

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Problem 6.1.12

If A\mathbf{A} is an n×nn \times n skew-symmetric matrix, prove that A\mathbf{A} is singular whenever nn is odd.

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Remark.
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Use Exercise 6.1.11.

Problem 6.1.13

How can you build random integer matrices with det⁡(A)=1\operatorname {det}\left(\mathbf{A}\right) = 1?

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Problem 6.1.14

If the kthk^{th} row of An×n\mathbf{A}_{n \times n} is written as a sum Ak∗=x→T+y→T+⋯+z→T\mathbf{A}_{k*} = \overrightarrow {x}^{T} + \overrightarrow {y}^{T} + \cdots + \overrightarrow {z}^{T}, where x→T,y→T,…,z→T\overrightarrow {x}^{T}, \overrightarrow {y}^{T}, \ldots , \overrightarrow {z}^{T} are row vectors, explain why

det⁡(A)=det⁡(A1∗⋮x→T⋮An∗)+det⁡(A1∗⋮y→T⋮An∗)+⋯+det⁡(A1∗⋮z→T⋮An∗). \operatorname {det}\left(\mathbf{A}\right) = \det \begin{pmatrix} \mathbf{A}_{1*} \\ \vdots \\ \overrightarrow {x}^{T} \\ \vdots \\ \mathbf{A}_{n*} \end{pmatrix} + \det \begin{pmatrix} \mathbf{A}_{1*} \\ \vdots \\ \overrightarrow {y}^{T} \\ \vdots \\ \mathbf{A}_{n*} \end{pmatrix} + \cdots + \det \begin{pmatrix} \mathbf{A}_{1*} \\ \vdots \\ \overrightarrow {z}^{T} \\ \vdots \\ \mathbf{A}_{n*} \end{pmatrix}.
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Problem 6.1.15

The CBS inequality (Cauchy--Bunyakovskii--Schwarz inequality) says that ∣x→∗y→∣≤∥x→∥2∥y→∥2\left|\overrightarrow {x}^{*}\overrightarrow {y}\right| \leq \left\| \overrightarrow {x}\right\|_{2} \left\| \overrightarrow {y}\right\|_{2} for vectors x→,y→∈Cn×1\overrightarrow {x}, \overrightarrow {y} \in \mathbb {C}^{n \times 1}, with equality if and only if y→=αx→\overrightarrow {y} = \alpha \overrightarrow {x} for α=x→∗y→/x→∗x→\alpha = \overrightarrow {x}^{*}\overrightarrow {y}/\overrightarrow {x}^{*}\overrightarrow {x}. Use Exercise 6.1.10 to give an alternate proof of the CBS inequality along with an alternate explanation of why equality holds if and only if y→\overrightarrow {y} is a scalar multiple of x→\overrightarrow {x}.

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Problem 6.1.16

Let Ak\mathbf{A}_{k} be the k×kk \times k leading principal submatrix of An×n\mathbf{A}_{n \times n} --- i.e., the submatrix taken from the upper-left-hand corner of A\mathbf{A} consisting of its first kk rows and columns. Prove that if A\mathbf{A} has an LU factorization A=LU\mathbf{A} = \mathbf{L}\mathbf{U}, then det⁡(Ak)=u11u22⋯ukk\operatorname {det}\left(\mathbf{A}_{k}\right) = u_{11}u_{22} \cdots u_{kk}, and deduce that the kthk^{th} pivot is

ukk={det⁡(A1)=a11for k=1,det⁡(Ak)/det⁡(Ak−1)for k=2,3,…,n. u_{kk} = \begin{cases} \operatorname {det}\left(\mathbf{A}_{1}\right) = a_{11} & \text{for } k = 1, \\ \operatorname {det}\left(\mathbf{A}_{k}\right)/\operatorname {det}\left(\mathbf{A}_{k-1}\right) & \text{for } k = 2,3,\ldots ,n. \end{cases}
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Problem 6.1.17

Prove that if rk⁡(Am×n)=n\operatorname {rk}\left(\mathbf{A}_{m \times n}\right) = n, then ATA\mathbf{A}^{T}\mathbf{A} has an LU factorization with positive pivots --- i.e., ATA\mathbf{A}^{T}\mathbf{A} is positive definite, meaning that the symmetric matrix ATA\mathbf{A}^{T}\mathbf{A} possesses an LU factorization in which every pivot is positive.

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Problem 6.1.18

Let A(x)=(2−x3404−x−51−13−x)\mathbf{A}(x) = \begin{pmatrix} 2-x & 3 & 4 \\ 0 & 4-x & -5 \\ 1 & -1 & 3-x \end{pmatrix}.

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(a)

First evaluate det⁡(A)\operatorname {det}\left(\mathbf{A}\right), and then compute d(det⁡(A))/dxd\bigl(\operatorname {det}\left(\mathbf{A}\right)\bigr)/dx.

(b)

Use formula (6.1.19) to evaluate d(det⁡(A))/dxd\bigl(\operatorname {det}\left(\mathbf{A}\right)\bigr)/dx.

Problem 6.1.19

When the entries of A=[aij(x)]\mathbf{A} = [a_{ij}(x)] are differentiable functions of xx, we define dA/dx=[daij/dx]d\mathbf{A}/dx = [da_{ij}/dx] (the matrix of derivatives). For square matrices, is it always the case that d(det⁡(A))/dx=det⁡(dA/dx)d\bigl(\operatorname {det}\left(\mathbf{A}\right)\bigr)/dx = \operatorname {det}\left(d\mathbf{A}/dx\right)?

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Problem 6.1.20

For a set of functions S={f1(x),f2(x),…,fn(x)}\mathcal{S} = \left\{ f_{1}(x), f_{2}(x), \ldots , f_{n}(x)\right\} that are n−1n-1 times differentiable, the determinant

w(x)=∣f1(x)f2(x)⋯fn(x)f1′(x)f2′(x)⋯fn′(x)⋮⋮⋱⋮f1(n−1)(x)f2(n−1)(x)⋯fn(n−1)(x)∣ w(x) = \begin{vmatrix} f_{1}(x) & f_{2}(x) & \cdots & f_{n}(x) \\ f_{1}'(x) & f_{2}'(x) & \cdots & f_{n}'(x) \\ \vdots & \vdots & \ddots & \vdots \\ f_{1}^{(n-1)}(x) & f_{2}^{(n-1)}(x) & \cdots & f_{n}^{(n-1)}(x) \end{vmatrix}

is called the Wronskian of S\mathcal{S}. If S\mathcal{S} is a linearly dependent set, explain why w(x)=0w(x) = 0 for every value of xx.

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Remark.
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Recall Example 4.3.6 (p. 189).

Problem 6.1.21

Consider evaluating an n×nn \times n determinant from the definition (6.1.1).

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(a)

How many multiplications are required?

(b)

Assuming a computer will do 1,000,000 multiplications per second, and neglecting all other operations, what is the largest order determinant that can be evaluated in one hour?

(c)

Under the same conditions of part (b), how long will it take to evaluate the determinant of a 100×100100 \times 100 matrix? Hint: 100!≈9.33×10157100! \approx 9.33 \times 10^{157}.

(d)

If all other operations are neglected, how many multiplications per second must a computer perform if the task of evaluating the determinant of a 100×100100 \times 100 matrix is to be completed in 100 years?