Chapter 6

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?

Problem 6.2.1

Use a cofactor expansion to evaluate each of the following determinants.

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(a)

∣211621−221∣\begin{vmatrix} 2 & 1 & 1 \\ 6 & 2 & 1 \\ -2 & 2 & 1 \end{vmatrix}.

(b)

∣00−231012−112102−30∣\begin{vmatrix} 0 & 0 & -2 & 3 \\ 1 & 0 & 1 & 2 \\ -1 & 1 & 2 & 1 \\ 0 & 2 & -3 & 0 \end{vmatrix}.

(c)

∣0111101111011110∣\begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{vmatrix}.

Problem 6.2.2

Use determinants to compute the following inverses.

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(a)

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

(b)

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

Problem 6.2.3
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(a)

Use Cramer's rule to solve

x1+x2+x3=1,x1+x2=α,x2+x3=β. \begin{aligned} x_{1} + x_{2} + x_{3} & = 1, \\ x_{1} + x_{2} & = \alpha , \\ x_{2} + x_{3} & = \beta . \end{aligned}
(b)

Evaluate lim⁡t→∞x2(t)\lim_{t \to \infty } x_{2}(t), where x2(t)x_{2}(t) is defined by the system

x1+tx2+t2x3=t4,t2x1+x2+tx3=t3,tx1+t2x2+x3=0. \begin{aligned} x_{1} + tx_{2} + t^{2}x_{3} & = t^{4}, \\ t^{2}x_{1} + x_{2} + tx_{3} & = t^{3}, \\ tx_{1} + t^{2}x_{2} + x_{3} & = 0. \end{aligned}
Problem 6.2.4

Is the following equation a valid derivation of Cramer's rule for solving a nonsingular system Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b}, where Ai\mathbf{A}_{i} denotes the matrix identical to A\mathbf{A} except that column A∗i\mathbf{A}_{*i} has been replaced by b→\overrightarrow {b}?

det⁡(Ai)det⁡(A)=det⁡(A−1Ai)=det⁡(e→1⋯e→i−1  x→  e→i+1⋯e→n)=xi. \dfrac {\operatorname {det}\left(\mathbf{A}_{i}\right)}{\operatorname {det}\left(\mathbf{A}\right)} = \operatorname {det}\left(\mathbf{A}^{-1}\mathbf{A}_{i}\right) = \operatorname {det}\left(\overrightarrow {e}_{1} \cdots \overrightarrow {e}_{i-1} \; \overrightarrow {x} \; \overrightarrow {e}_{i+1} \cdots \overrightarrow {e}_{n}\right) = x_{i}.
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Problem 6.2.5
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(a)

By example, show that det⁡(A+B)≠det⁡(A)+det⁡(B)\operatorname {det}\left(\mathbf{A}+\mathbf{B}\right) \neq \operatorname {det}\left(\mathbf{A}\right)+\operatorname {det}\left(\mathbf{B}\right).

(b)

Using square matrices, construct an example that shows that

det⁡(ABCD)≠det⁡(A)det⁡(D)−det⁡(B)det⁡(C). \det \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} \neq \operatorname {det}\left(\mathbf{A}\right)\operatorname {det}\left(\mathbf{D}\right) - \operatorname {det}\left(\mathbf{B}\right)\operatorname {det}\left(\mathbf{C}\right).
Problem 6.2.6

Suppose rk⁡(Bm×n)=n\operatorname {rk}\left(\mathbf{B}_{m \times n}\right) = n, and let Q\mathbf{Q} be the orthogonal projector onto N(BT)N(\mathbf{B}^{T}) --- i.e., Q=I−B(BTB)−1BT\mathbf{Q} = \mathbf{I} - \mathbf{B}(\mathbf{B}^{T}\mathbf{B})^{-1}\mathbf{B}^{T}, the projector that annihilates R(B)R(\mathbf{B}) and reproduces vectors already in N(BT)N(\mathbf{B}^{T}). For A=[B∣c→n×1]\mathbf{A} = [\mathbf{B} \mid \overrightarrow {c}_{n \times 1}], prove c→TQc→=det⁡(ATA)/det⁡(BTB)\overrightarrow {c}^{T}\mathbf{Q}\overrightarrow {c} = \operatorname {det}\left(\mathbf{A}^{T}\mathbf{A}\right)/\operatorname {det}\left(\mathbf{B}^{T}\mathbf{B}\right).

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Problem 6.2.7

If An×n\mathbf{A}_{n \times n} is a nonsingular matrix, and if D\mathbf{D} and C\mathbf{C} are n×kn \times k matrices, explain how to use (6.2.1) to derive the formula

det⁡(A+CDT)=det⁡(A)det⁡(Ik+DTA−1C). \operatorname {det}\left(\mathbf{A}+\mathbf{C}\mathbf{D}^{T}\right) = \operatorname {det}\left(\mathbf{A}\right)\operatorname {det}\left(\mathbf{I}_{k}+\mathbf{D}^{T}\mathbf{A}^{-1}\mathbf{C}\right).
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Remark.
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This is a generalization of the rank-one update formula det⁡(A+c→d→T)=det⁡(A)(1+d→TA−1c→)\operatorname {det}\left(\mathbf{A}+\overrightarrow {c}\overrightarrow {d}^{T}\right) = \operatorname {det}\left(\mathbf{A}\right)\bigl(1+\overrightarrow {d}^{T}\mathbf{A}^{-1}\overrightarrow {c}\bigr) (6.2.3) because if c→i\overrightarrow {c}_{i} and d→i\overrightarrow {d}_{i} are the ithi^{th} columns of C\mathbf{C} and D\mathbf{D}, respectively, then A+CDT=A+c→1d→1T+c→2d→2T+⋯+c→kd→kT\mathbf{A}+\mathbf{C}\mathbf{D}^{T} = \mathbf{A}+\overrightarrow {c}_{1}\overrightarrow {d}_{1}^{T}+\overrightarrow {c}_{2}\overrightarrow {d}_{2}^{T}+\cdots +\overrightarrow {c}_{k}\overrightarrow {d}_{k}^{T}.

Problem 6.2.8

Explain why A\mathbf{A} is singular if and only if A[adj⁡(A)]=0\mathbf{A}[\operatorname {adj}(\mathbf{A})] = \mathbf{0}.

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Problem 6.2.9

For a nonsingular linear system Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b}, explain why each component of the solution must vary continuously with the entries of A\mathbf{A}.

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Problem 6.2.10

For scalars α\alpha, explain why adj⁡(αA)=αn−1adj⁡(A)\operatorname {adj}(\alpha \mathbf{A}) = \alpha^{n-1}\operatorname {adj}(\mathbf{A}).

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Remark.
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Recall Exercise 6.1.11.

Problem 6.2.11

For an n×nn \times n matrix A\mathbf{A}, prove that the following statements are true.

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(a)

If rk⁡(A)<n−1\operatorname {rk}\left(\mathbf{A}\right) < n-1, then adj⁡(A)=0\operatorname {adj}(\mathbf{A}) = \mathbf{0}.

(b)

If rk⁡(A)=n−1\operatorname {rk}\left(\mathbf{A}\right) = n-1, then rk⁡(adj⁡(A))=1\operatorname {rk}\left(\operatorname {adj}(\mathbf{A})\right) = 1.

(c)

If rk⁡(A)=n\operatorname {rk}\left(\mathbf{A}\right) = n, then rk⁡(adj⁡(A))=n\operatorname {rk}\left(\operatorname {adj}(\mathbf{A})\right) = n.

Problem 6.2.12

In 1812, Cauchy discovered the formula that says that if A\mathbf{A} is n×nn \times n, then det⁡(adj⁡(A))=[det⁡(A)]n−1\operatorname {det}\left(\operatorname {adj}(\mathbf{A})\right) = [\operatorname {det}\left(\mathbf{A}\right)]^{n-1}. Establish Cauchy's formula.

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Problem 6.2.13

For the following tridiagonal matrix, An\mathbf{A}_{n}, let Dn=det⁡(An)D_{n} = \operatorname {det}\left(\mathbf{A}_{n}\right), and derive the formula Dn=2Dn−1−Dn−2D_{n} = 2D_{n-1}-D_{n-2} to deduce that Dn=n+1D_{n} = n+1.

An=(2−10⋯0−12−1⋯0⋱⋱⋱0⋯−12−10⋯0−12)n×n. \mathbf{A}_{n} = \begin{pmatrix} 2 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & \cdots & 0 \\ & \ddots & \ddots & \ddots & \\ 0 & \cdots & -1 & 2 & -1 \\ 0 & \cdots & 0 & -1 & 2 \end{pmatrix}_{n \times n}.
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Problem 6.2.14

By considering rank-one updated matrices, derive the following formulas.

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(a)

∣1+α1α11⋯111+α2α2⋯1⋮⋮⋱⋮11⋯1+αnαn∣=1+∑αi∏αi\begin{vmatrix} \frac{1+\alpha_{1}}{\alpha_{1}} & 1 & \cdots & 1 \\ 1 & \frac{1+\alpha_{2}}{\alpha_{2}} & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & \frac{1+\alpha_{n}}{\alpha_{n}} \end{vmatrix} = \dfrac {1+\sum \alpha_{i}}{\prod \alpha_{i}}.

(b)

∣αββ⋯ββαβ⋯βββα⋯β⋮⋮⋮⋱⋮βββ⋯α∣n×n={(α−β)n(1+nβα−β)if α≠β,0if α=β.\begin{vmatrix} \alpha & \beta & \beta & \cdots & \beta \\ \beta & \alpha & \beta & \cdots & \beta \\ \beta & \beta & \alpha & \cdots & \beta \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \beta & \beta & \beta & \cdots & \alpha \end{vmatrix}_{n \times n} = \begin{cases} (\alpha -\beta )^{n}\left(1+\dfrac {n\beta }{\alpha -\beta }\right) & \text{if } \alpha \neq \beta , \\ 0 & \text{if } \alpha = \beta . \end{cases}

(c)

∣1+α1α2⋯αnα11+α2⋯αn⋮⋮⋱⋮α1α2⋯1+αn∣=1+α1+α2+⋯+αn\begin{vmatrix} 1+\alpha_{1} & \alpha_{2} & \cdots & \alpha_{n} \\ \alpha_{1} & 1+\alpha_{2} & \cdots & \alpha_{n} \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{1} & \alpha_{2} & \cdots & 1+\alpha_{n} \end{vmatrix} = 1+\alpha_{1}+\alpha_{2}+\cdots +\alpha_{n}.

Problem 6.2.15

A bordered matrix has the form B=(Ax→y→Tα)\mathbf{B} = \begin{pmatrix} \mathbf{A} & \overrightarrow {x} \\ \overrightarrow {y}^{T} & \alpha \end{pmatrix} in which An×n\mathbf{A}_{n \times n} is nonsingular, x→\overrightarrow {x} is a column, y→T\overrightarrow {y}^{T} is a row, and α\alpha is a scalar. Explain why the following statements must be true.

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(a)

∣Ax→y→T−1∣=−det⁡(A+x→y→T)\begin{vmatrix} \mathbf{A} & \overrightarrow {x} \\ \overrightarrow {y}^{T} & -1 \end{vmatrix} = -\operatorname {det}\left(\mathbf{A}+\overrightarrow {x}\overrightarrow {y}^{T}\right).

(b)

∣Ax→y→T0∣=−y→Tadj⁡(A)x→\begin{vmatrix} \mathbf{A} & \overrightarrow {x} \\ \overrightarrow {y}^{T} & 0 \end{vmatrix} = -\overrightarrow {y}^{T}\operatorname {adj}(\mathbf{A})\overrightarrow {x}.

Problem 6.2.16

If B\mathbf{B} is m×nm \times n and C\mathbf{C} is n×mn \times m, explain why (6.2.1) guarantees that λmdet⁡(λIn−CB)=λndet⁡(λIm−BC)\lambda^{m}\operatorname {det}\left(\lambda \mathbf{I}_{n}-\mathbf{C}\mathbf{B}\right) = \lambda^{n}\operatorname {det}\left(\lambda \mathbf{I}_{m}-\mathbf{B}\mathbf{C}\right) is true for all scalars λ\lambda.

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Problem 6.2.17

For a square matrix A\mathbf{A} and column vectors c→\overrightarrow {c} and d→\overrightarrow {d}, derive the following two extensions of the rank-one update formula det⁡(A+c→d→T)=det⁡(A)(1+d→TA−1c→)\operatorname {det}\left(\mathbf{A}+\overrightarrow {c}\overrightarrow {d}^{T}\right) = \operatorname {det}\left(\mathbf{A}\right)\bigl(1+\overrightarrow {d}^{T}\mathbf{A}^{-1}\overrightarrow {c}\bigr) (6.2.3).

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(a)

If Ax→=c→\mathbf{A}\overrightarrow {x} = \overrightarrow {c}, then det⁡(A+c→d→T)=det⁡(A)(1+d→Tx→)\operatorname {det}\left(\mathbf{A}+\overrightarrow {c}\overrightarrow {d}^{T}\right) = \operatorname {det}\left(\mathbf{A}\right)\bigl(1+\overrightarrow {d}^{T}\overrightarrow {x}\bigr).

(b)

If y→TA=d→T\overrightarrow {y}^{T}\mathbf{A} = \overrightarrow {d}^{T}, then det⁡(A+c→d→T)=det⁡(A)(1+y→Tc→)\operatorname {det}\left(\mathbf{A}+\overrightarrow {c}\overrightarrow {d}^{T}\right) = \operatorname {det}\left(\mathbf{A}\right)\bigl(1+\overrightarrow {y}^{T}\overrightarrow {c}\bigr).

Problem 6.2.18

Describe the determinant of an elementary reflector R=I−2u→u→∗/u→∗u→\mathbf{R} = \mathbf{I} - 2\overrightarrow {u}\overrightarrow {u}^{*}/\overrightarrow {u}^{*}\overrightarrow {u} (a reflection about the hyperplane u→⊥\overrightarrow {u}^{\perp }) and a plane rotation Pij\mathbf{P}_{ij} (a rotation in the (i,j)(i,j)-plane, using cos⁡θ\cos \theta and sin⁡θ\sin \theta in positions (i,i),(i,j),(j,i),(j,j)(i,i),(i,j),(j,i),(j,j) of an otherwise identity matrix), and then explain how to find det⁡(A)\operatorname {det}\left(\mathbf{A}\right) using Householder reduction (reducing A\mathbf{A} to upper-trapezoidal form by a sequence of elementary reflectors) and Givens reduction (the same reduction using a sequence of plane rotations instead).

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Problem 6.2.19

Suppose that A\mathbf{A} is a nonsingular matrix whose entries are integers. Prove that the entries in A−1\mathbf{A}^{-1} are integers if and only if det⁡(A)=±1\operatorname {det}\left(\mathbf{A}\right) = \pm 1.

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Problem 6.2.20

Let A=I−2u→v→T\mathbf{A} = \mathbf{I} - 2\overrightarrow {u}\overrightarrow {v}^{T} be a matrix in which u→\overrightarrow {u} and v→\overrightarrow {v} are column vectors with integer entries.

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(a)

Prove that A−1\mathbf{A}^{-1} has integer entries if and only if v→Tu→=0\overrightarrow {v}^{T}\overrightarrow {u} = 0 or 11.

(b)

A matrix is said to be involutory whenever A−1=A\mathbf{A}^{-1} = \mathbf{A}. Explain why A=I−2u→v→T\mathbf{A} = \mathbf{I}-2\overrightarrow {u}\overrightarrow {v}^{T} is involutory when v→Tu→=1\overrightarrow {v}^{T}\overrightarrow {u} = 1.

Problem 6.2.21

Use induction to argue that a cofactor expansion of det⁡(An×n)\operatorname {det}\left(\mathbf{A}_{n \times n}\right) requires

c(n)=n! (1+12!+13!+⋯+1(n−1)!) c(n) = n!\, \left(1+\dfrac {1}{2!}+\dfrac {1}{3!}+\cdots +\dfrac {1}{(n-1)!}\right)

multiplications for n≥2n \geq 2. Assume a computer will do 1,000,000 multiplications per second, and neglect all other operations to estimate how long it will take to evaluate the determinant of a 100×100100 \times 100 matrix using cofactor expansions.

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Remark.
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Recall the series expansion for exe^{x}, and use 100!≈9.33×10157100! \approx 9.33 \times 10^{157}.

Problem 6.2.22

Determine all values of λ\lambda for which the matrix A−λI\mathbf{A}-\lambda \mathbf{I} is singular, where

A=(0−3−2252−2−30). \mathbf{A} = \begin{pmatrix} 0 & -3 & -2 \\ 2 & 5 & 2 \\ -2 & -3 & 0 \end{pmatrix}.
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Remark.
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If p(λ)=λn+αn−1λn−1+⋯+α1λ+α0p(\lambda ) = \lambda^{n}+\alpha_{n-1}\lambda^{n-1}+\cdots +\alpha_{1}\lambda +\alpha_{0} is a monic polynomial with integer coefficients, then the integer roots of p(λ)p(\lambda ) are a subset of the factors of α0\alpha_{0}.

Problem 6.2.23

Suppose that f1(t),f2(t),…,fn(t)f_{1}(t), f_{2}(t), \ldots , f_{n}(t) are solutions of the nthn^{th}-order linear differential equation y(n)+p1(t)y(n−1)+⋯+pn−1(t)y′+pn(t)y=0y^{(n)}+p_{1}(t)y^{(n-1)}+\cdots +p_{n-1}(t)y'+p_{n}(t)y = 0, and let w(t)w(t) be the Wronskian

w(t)=∣f1(t)f2(t)⋯fn(t)f1′(t)f2′(t)⋯fn′(t)⋮⋮⋱⋮f1(n−1)(t)f2(n−1)(t)⋯fn(n−1)(t)∣. w(t) = \begin{vmatrix} f_{1}(t) & f_{2}(t) & \cdots & f_{n}(t) \\ f_{1}'(t) & f_{2}'(t) & \cdots & f_{n}'(t) \\ \vdots & \vdots & \ddots & \vdots \\ f_{1}^{(n-1)}(t) & f_{2}^{(n-1)}(t) & \cdots & f_{n}^{(n-1)}(t) \end{vmatrix}.

By converting the nthn^{th}-order equation into a system of nn first-order equations with the substitutions x1=y,x2=y′,…,xn=y(n−1)x_{1} = y, x_{2} = y', \ldots , x_{n} = y^{(n-1)}, show that w(t)=w(ξ0) e−∫ξ0tp1(ξ) dξw(t) = w(\xi_{0})\, e^{-\int_{\xi_{0}}^{t} p_{1}(\xi )\, d\xi } for an arbitrary constant ξ0\xi_{0}.

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Problem 6.2.24

Evaluate the Vandermonde determinant by showing

∣1x1x12⋯x1n−11x2x22⋯x2n−1⋮⋮⋮⋯⋮1xnxn2⋯xnn−1∣=∏j>i(xj−xi). \begin{vmatrix} 1 & x_{1} & x_{1}^{2} & \cdots & x_{1}^{n-1} \\ 1 & x_{2} & x_{2}^{2} & \cdots & x_{2}^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & x_{n} & x_{n}^{2} & \cdots & x_{n}^{n-1} \end{vmatrix} = \prod _{j>i} (x_{j}-x_{i}).

When is this nonzero (compare with Example 4.3.4)?

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Remark.
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For the polynomial p(λ)=∣1λλ2⋯λk−11x2x22⋯x2k−1⋮⋮⋮⋯⋮1xkxk2⋯xkk−1∣k×kp(\lambda ) = \begin{vmatrix} 1 & \lambda & \lambda^{2} & \cdots & \lambda^{k-1} \\ 1 & x_{2} & x_{2}^{2} & \cdots & x_{2}^{k-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & x_{k} & x_{k}^{2} & \cdots & x_{k}^{k-1} \end{vmatrix}_{k \times k}, use induction to find the degree of p(λ)p(\lambda ), the roots of p(λ)p(\lambda ), and the coefficient of λk−1\lambda^{k-1} in p(λ)p(\lambda ).

Problem 6.2.25

Suppose that each entry in An×n=[aij(x)]\mathbf{A}_{n \times n} = [a_{ij}(x)] is a differentiable function of a real variable xx. Use formula (6.1.19) to derive the formula

d(det⁡(A))dx=∑j=1n∑i=1ndaijdx A˚ij. \dfrac {d\bigl(\operatorname {det}\left(\mathbf{A}\right)\bigr)}{dx} = \sum _{j=1}^{n} \sum _{i=1}^{n} \dfrac {da_{ij}}{dx}\, \mathring {A}_{ij}.
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Problem 6.2.26

Consider the entries of A\mathbf{A} to be independent variables, and use formula (6.1.19) to derive the formula

∂det⁡(A)∂aij=A˚ij. \dfrac {\partial \operatorname {det}\left(\mathbf{A}\right)}{\partial a_{ij}} = \mathring {A}_{ij}.
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Problem 6.2.27

In 1772, the French mathematician Pierre-Simon Laplace (1749--1827) presented the following generalized version of the cofactor expansion. For an n×nn \times n matrix A\mathbf{A}, let A(i1i2⋯ik∣j1j2⋯jk)\mathbf{A}(i_{1}i_{2} \cdots i_{k} \mid j_{1}j_{2} \cdots j_{k}) denote the k×kk \times k submatrix of A\mathbf{A} that lies on the intersection of rows i1,i2,…,iki_{1},i_{2},\ldots ,i_{k} with columns j1,j2,…,jkj_{1},j_{2},\ldots ,j_{k}, and let M(i1i2⋯ik∣j1j2⋯jk)M(i_{1}i_{2} \cdots i_{k} \mid j_{1}j_{2} \cdots j_{k}) denote the n−k×n−kn-k \times n-k minor determinant obtained by deleting rows i1,i2,…,iki_{1},i_{2},\ldots ,i_{k} and columns j1,j2,…,jkj_{1},j_{2},\ldots ,j_{k} from A\mathbf{A}. The generalized cofactor of A(i1⋯ik∣j1⋯jk)\mathbf{A}(i_{1} \cdots i_{k} \mid j_{1} \cdots j_{k}) is defined to be the signed minor A˚(i1⋯ik∣j1⋯jk)=(−1)i1+⋯+ik+j1+⋯+jkM(i1⋯ik∣j1⋯jk)\mathring {A}(i_{1} \cdots i_{k} \mid j_{1} \cdots j_{k}) = (-1)^{i_{1}+\cdots +i_{k}+j_{1}+\cdots +j_{k}}M(i_{1} \cdots i_{k} \mid j_{1} \cdots j_{k}), and for each fixed set of row indices 1≤i1<⋯<ik≤n1 \leq i_{1} < \cdots < i_{k} \leq n,

det⁡(A)=∑1≤j1<⋯<jk≤ndet⁡A(i1⋯ik∣j1⋯jk) A˚(i1⋯ik∣j1⋯jk). \operatorname {det}\left(\mathbf{A}\right) = \sum _{1 \leq j_{1} < \cdots < j_{k} \leq n} \det \mathbf{A}(i_{1} \cdots i_{k} \mid j_{1} \cdots j_{k})\, \mathring {A}(i_{1} \cdots i_{k} \mid j_{1} \cdots j_{k}).

Each such sum contains (nk)\binom {n}{k} terms. Use Laplace's expansion to evaluate the determinant of

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}

in terms of the first and third rows.

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