Determinants; Systems of Linear Equations

Excerpts

Determinants; Systems of Linear Equations
If $D \neq 0$, the unique values of $x_1, \dotsc, x_n$ determined by division from $(13)$ actually satisfy equations $(12)$.
Determinants; Systems of Linear Equations
If $D$ denotes the determinant of the coefficients of the $n$ unknowns in a system of $n$ linear equations, the product of $D$ by any one of the unknowns is equal to the determinant obtained from $D$ by substituting the known terms in place of the coefficients of that unknown. If $D \neq 0$, we obtain the unique values of the unknowns by division by $D$.
Determinants; Systems of Linear Equations
Multiply the members of the first equation by $b_2$ and those of the second equation by $-b_1$, and add the resulting equations.
Determinants; Systems of Linear Equations
*if $D$ is the determinant of the coefficients of the unknowns, the product of $D$ by any one of the unknowns is equal to the determinant obtained from $D$ by substituting the known terms in place of the coefficients of that unknown*.
Determinants; Systems of Linear Equations
The nine numbers $a_1, \dotsc, c_3$ are called the *elements* of the determinant.
Determinants; Systems of Linear Equations
If $D \ne 0$, relations $(3)$ uniquely determine values of $x$ and $y$:
Determinants; Systems of Linear Equations
A system of $m$ linear equations in $n$ unknowns is consistent if and only if the rank of the matrix of the coefficients of the unknowns is equal to the rank of the augmented matrix.
Determinants; Systems of Linear Equations
For example, a determinant $D$ of order $3$ is of rank $3$ if $D \neq 0$; of rank $2$ if $D = 0$, but some two-rowed minor is not zero; of rank $1$ if every two-rowed minor is zero, but some element is not zero.
Determinants; Systems of Linear Equations
A necessary and sufficient condition that $n$ linear homogeneous equations in $n$ unknowns shall have a set of solutions, other than the trivial one in which each unknown is zero, is that the determinant of the coefficients be zero.
Determinants; Systems of Linear Equations
For $r = 1$, this development becomes the known expansion of $D$ according to the elements of the first column (§90); here $M_1 = e_{11}$.
Determinants; Systems of Linear Equations
The theorem was discovered by induction in 1750 by G. Cramer.

Equations

Determinants; Systems of Linear Equations
a_{11} x_1 + a_{12} x_2 + \dotsb + a_{1n} x_n = k_1
The first of n linear equations in n unknowns, whose left side is a weighted sum of the unknowns and whose right side is a known term.
Determinants; Systems of Linear Equations
D = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \Dots{4} \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}
D is defined as the determinant of the coefficients of the n unknowns.
Determinants; Systems of Linear Equations
Dx_1 = K_1,\qquad Dx_2 = K_2,\qquad \dotsc,\qquad Dx_n = K_n
The product of D with each unknown equals the determinant K_i obtained from D by replacing the i-th column of coefficients with the known terms, so each unknown is K_i divided by D when D is nonzero.
Determinants; Systems of Linear Equations
K_1 = \begin{vmatrix} k_1 & a_{12} & \cdots & a_{1n} \\ \Dots{4} \\ k_n & a_{n2} & \cdots & a_{nn} \end{vmatrix}
K_1 is the determinant obtained from D by replacing the first column of coefficients with the known terms.
Determinants; Systems of Linear Equations
L_i \equiv a_{i1} x_1 + a_{i2} x_2 + \dotsb + a_{in} x_n - k_i
L_i is the i-th equation with its known term transposed to the left, so that it is zero when the equation holds.
Determinants; Systems of Linear Equations
K = \begin{vmatrix} a_{11} & \cdots & a_{1r} & k_1 \\ \Dots{4} \\ a_{r+11} & \cdots & a_{r+1r} & k_{r+1} \end{vmatrix}
K is the determinant formed from the first r+1 rows by putting the known terms in the last column.
Determinants; Systems of Linear Equations
0 = ±K
Multiplying and adding the first r+1 equations with the minors of the known terms gives the identity 0 = ±K, which forces K to vanish if the equations are consistent.
Determinants; Systems of Linear Equations
d_{r+1} = \begin{vmatrix} a_{11} & \cdots & a_{1r} \\ \Dots{3} \\ a_{r1} & \cdots & a_{rr} \end{vmatrix}
d_{r+1} is the nonvanishing r-rowed minor formed from the first r rows and first r columns of the coefficient determinant.
Determinants; Systems of Linear Equations
d_1L_1 - d_2L_2 + \dotsb + (-1)^rd_{r+1}L_{r+1} = \mp K = 0
A signed combination of the first r+1 equations, with minors as coefficients, reduces to zero, so one equation is a linear combination of the others.
Determinants; Systems of Linear Equations
A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \Dots{4} \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}
A is the matrix of the coefficients of the unknowns, arranged as they occur in the equations.
Determinants; Systems of Linear Equations
B = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & k_1\\ \Dots{5}\\ a_{m1} & a_{m2} & \cdots & a_{mn} & k_m \end{pmatrix}
B is the matrix of coefficients with the column of known terms annexed.
Determinants; Systems of Linear Equations
D = \begin{vmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{vmatrix}
D is a determinant of order 4 whose elements are the letters a_i, b_i, c_i, d_i, used as the example for complementary minors.
Determinants; Systems of Linear Equations
M = \begin{vmatrix} a_1 & b_1 \\ a_3 & b_3 \end{vmatrix}
M is the two-rowed minor formed from rows 1 and 3 and columns 1 and 2 of D.
Determinants; Systems of Linear Equations
M' = \begin{vmatrix} c_2 & d_2 \\ c_4 & d_4 \end{vmatrix}
M' is the two-rowed minor complementary to M, formed from rows 2 and 4 and columns 3 and 4 of D.
Determinants; Systems of Linear Equations
\begin{vmatrix} a & b \\ c & d \end{vmatrix} · \begin{vmatrix} e & f \\ g & h \end{vmatrix} = \begin{vmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{vmatrix}
The product of two determinants of order 2 equals the determinant whose elements are the row-by-column sums of products.
Determinants; Systems of Linear Equations
\begin{vmatrix} \Neg a_1 & \Neg b_1 & \Neg c_1 & 0 & 0 & 0 \\ \Neg a_2 & \Neg b_2 & \Neg c_2 & 0 & 0 & 0 \\ \Neg a_3 & \Neg b_3 & \Neg c_3 & 0 & 0 & 0 \\ -1 & \Neg 0 & \Neg 0 & e_1 & f_1 & g_1 \\ \Neg 0 & -1 & \Neg 0 & e_2 & f_2 & g_2 \\ \Neg 0 & \Neg 0 & -1 & e_3 & f_3 & g_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} · \begin{vmatrix} e_1 & f_1 & g_1 \\ e_2 & f_2 & g_2 \\ e_3 & f_3 & g_3 \end{vmatrix}
The product of two third-order determinants equals a determinant of order 6, obtained by a Laplace development with r = 3.
Determinants; Systems of Linear Equations
x_1 = 0, \dotsc, x_n = 0
Every unknown equal to zero satisfies any set of homogeneous linear equations, so this is always a solution.
Determinants; Systems of Linear Equations
D = \sum_{j=1}^n (-1)^{j+k} e_{jk} E_{jk}
Any determinant of order n equals the signed sum of the elements of its kth column times their minors (expansion according to any column).
Determinants; Systems of Linear Equations
a_1 b_2 - a_2 b_1
The determinant of the second order is defined as the common multiplier a_1 b_2 - a_2 b_1 of x and y in the two-equation system.
Determinants; Systems of Linear Equations
x = \frac{k_1 b_2 - k_2 b_1}{D}
When D is not zero, the unknown x equals the determinant with the known terms substituted in the first column, divided by D.
Determinants; Systems of Linear Equations
y = \frac{a_1 k_2 - a_2 k_1}{D}
When D is not zero, the unknown y equals the determinant with the known terms substituted in the second column, divided by D.
Determinants; Systems of Linear Equations
a_1 b_2 c_3 - a_1 b_3 c_2 + a_2 b_3 c_1 - a_2 b_1 c_3 + a_3 b_1 c_2 - a_3 b_2 c_1
The determinant of the third order is defined as the signed sum of the six products taking one element from each row and column, with signs fixed by the parity of the subscript arrangement.
Determinants; Systems of Linear Equations
\sum_{(24)} ± a_q b_r c_s d_t
A determinant of order 4 is the sum over all 24 arrangements q, r, s, t of 1, 2, 3, 4 of the products a_q b_r c_s d_t, each with the sign + or - according to the parity of the arrangement.
Determinants; Systems of Linear Equations
(-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_n}n}
A term of a determinant of order n is a product with one element from each row and column, with sign (-1) to the power i, where i is the number of interchanges that produce the arrangement i_1, ..., i_n.
Determinants; Systems of Linear Equations
D = e_{11}E_{11} - e_{21}E_{21} + e_{31}E_{31} - \dotsb + (-1)^{n-1} e_{n1}E_{n1}
Any determinant of order n equals the alternating sum of its first-column elements times their minors (expansion according to the first column).
Determinants; Systems of Linear Equations
D = -a_2A_2 + b_2B_2 - c_2C_2
A third-order determinant equals the expansion along its second row, with alternating signs applied to each element times its minor.
Determinants; Systems of Linear Equations
D = -b_1B_1 + b_2B_2 - b_3B_3
A third-order determinant equals the expansion along its second column, with alternating signs applied to each element times its minor.
Determinants; Systems of Linear Equations
(-1)^m P \equiv (-1)^t P
If one arrangement is reached from 1, 2, ..., n by m interchanges and also by t interchanges, then m and t are both even or both odd.
Determinants; Systems of Linear Equations
\Delta = -D
Interchanging two columns of a determinant changes its sign, so the new determinant Delta equals minus D.
Determinants; Systems of Linear Equations
D = 0
A determinant is zero if any two of its rows or any two of its columns are alike.

Problems

Exercise Page115

Exercise Page115, problem 1, p. 115
$\begin{System}{3} x &+{}& y &+{}& z &= 11, \\ 2x &-{}& 6y &-{}& z &= 0, \\ 3x &+{}& 4y &+{}& 2z &= 0. \end{System}$
Printed answer:
- $x = -8$, $y = -7$, $z = 26$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: -8, y: -7, z: 26}
Exercise Page115, problem 2, p. 115
$\begin{System}{3} x &+{}& y &+{}& z &= 0, \\ x &+{}& 2y &+{}& 3z &= -1, \\ x &+{}& 3y &+{}& 6z &= 0. \end{System}$
Printed answer:
- $x = 3$, $y = -5$, $z = 2$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 3, y: -5, z: 2}
Exercise Page115, problem 3, p. 115
$\begin{System}{3} x &-{}& 2y &+{}& z &= 12, \\ x &+{}& 2y &+{}& 3z &= 48, \\ 6x &+{}& 4y &+{}& 3z &= 84. \end{System}$
Printed answer:
- $x = 6$, $y = 3$, $z = 12$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 6, y: 3, z: 12}
Exercise Page115, problem 4, p. 115
$\begin{System}{2} 3x &-{}& 2y &= 7, \\ 3y &-{}& 2z &= 6, \\ 3z &-{}& 2x &= -1. \end{System}$
Printed answer:
- $x = 5$, $y = 4$, $z = 3$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 5, y: 4, z: 3}
Exercise Page115, problem 5, p. 115
$\begin{System}{4} x &+{}& y &+{}& z &+{}& w &= 1, \\ x &+{}& 2y &+{}& 3z &+{}& 4w &= 11, \\ x &+{}& 3y &+{}& 6z &+{}& 10w &= 26, \\ x &+{}& 4y &+{}& 10z &+{}& 20w &= 47. \end{System}$
Printed answer:
- $x = -5$, $y = 3$, $z = 2$, $w = 1$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: -5, y: 3, z: 2, w: 1}
Exercise Page115, problem 6, p. 115
$\begin{System}{4} 2x &-{}& y &+{}& 3z &-{}& 2w &= 4, \\ x &+{}& 7y &+{}& z &-{}& w &= 2, \\ 3x &+{}& 5y &-{}& 5z &+{}& 3w &= 0, \\ 4x &-{}& 3y &+{}& 2z &-{}& w &= 5. \end{System}$
Printed answer:
- $x = 1$, $y = z = 0$, $w = -1$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 1, y: 0, z: 0, w: -1}
Exercise Page115, problem 7, p. 115
Prove the first relation $(13)$ by multiplying the members of the first equation $(12)$ by $A_{11}$, those of the second equation by $-A_{21}, \dotsc$, those of the $n$th equation by $(-1)^{n-1}A_{n1}$, and adding, where $A_{ij}$ by denotes the minor of $a_{ij}$ in $D$. Hint: The resulting coefficient of $x_2$ is the expansion, according to the elements of its first column, of a determinant derived from $D$ by replacing $a_{11}$ by $a_{12}$, $\dotsc$, $a_{n1}$ by $a_{n2}$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page119

Exercise Page119, problem 1, p. 119
$\begin{System}{3} 2x&+{}& y&+{}& 3z &= 1, \\ 4x&+{}& 2y&-{}& z &= -3, \\ 2x&+{}& y&-{}& 4z &= -4. \end{System}$
Printed answer:
- Consistent: $y = -8/7 - 2x$, $z = 5/7$ (common line).
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 2, p. 119
$\begin{System}{3} 2x&+{}& y&+{}& 3z &= 1, \\ 4x&+{}& 2y&-{}& z &= 3, \\ 2x&+{}& y&-{}& 4z &= 4. \end{System}$
Printed answer:
- Inconsistent, case $(\beta)$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 3, p. 119
$\begin{System}{3} x&-{}& 3y&+{}& 4z &= 1, \\ 4x&-{}& 12y&+{}& 16z &= 3, \\ 3x&-{}& 9y&+{}& 12z &= 3. \end{System}$
Printed answer:
- Inconsistent (two parallel planes).
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 4, p. 119
$\begin{System}{3} x&-{}& 3y&+{}& 4z &= 1, \\ 4x&-{}& 12y&+{}& 16z &= 4, \\ 3x&-{}& 9y&+{}& 12z &= 3. \end{System}$
Printed answer:
- Consistent (single plane).
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 5a, p. 119
Discuss the system System3 ax&+& y&+& z &= a-3, x&+& ay&+& z &= -2, x&+& y&+& az &= -2, System when (*i*) $a = 1$; (*ii*) $a = -2$; (*iii*) $a \neq 1$, $-2$, obtaining the simplest forms of the unknowns.
Printed answer:
- $z = -x-y-2$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 5b, p. 119
Discuss the system System3 ax&+& y&+& z &= a-3, x&+& ay&+& z &= -2, x&+& y&+& az &= -2, System when (*i*) $a = 1$; (*ii*) $a = -2$; (*iii*) $a \neq 1$, $-2$, obtaining the simplest forms of the unknowns.
Printed answer:
- inconsistent.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 5c, p. 119
Discuss the system System3 ax&+& y&+& z &= a-3, x&+& ay&+& z &= -2, x&+& y&+& az &= -2, System when (*i*) $a = 1$; (*ii*) $a = -2$; (*iii*) $a \neq 1$, $-2$, obtaining the simplest forms of the unknowns.
Printed answer:
- $x = \dfrac{a - 1}{a + 2}$, $y = z =\dfrac{-3}{a + 2}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: (a - 1)/(a + 2), y: -3/(a + 2), z: -3/(a + 2)}
Exercise Page119, problem 6a, p. 119
Discuss the system System3 x &+& y&+& z &= 1, ax &+& by&+& cz &= k, a^2x&+& b^2y&+& c^2z &= k^2, System when (*i*) $a$, $b$, $c$ are distinct; (*ii*) $a = b \neq c$; (*iii*) $a = b = c$.
Printed answer:
- $x = \dfrac{(k-b)(c-k)}{(a-b)(c-a)}$.
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem {x: (k-b)*(c-k)/((a-b)*(c-a))}
Exercise Page119, problem 6b, p. 119
Discuss the system System3 x &+& y&+& z &= 1, ax &+& by&+& cz &= k, a^2x&+& b^2y&+& c^2z &= k^2, System when (*i*) $a$, $b$, $c$ are distinct; (*ii*) $a = b \neq c$; (*iii*) $a = b = c$.
Printed answer:
- $y = \dfrac{k-c}{a-c}-x$, $z = \dfrac{a-k}{a-c}$ if $k=a$ or $k=c$, but inconsistent if $k$ is different from $a$ and $c$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page119, problem 6c, p. 119
Discuss the system System3 x &+& y&+& z &= 1, ax &+& by&+& cz &= k, a^2x&+& b^2y&+& c^2z &= k^2, System when (*i*) $a$, $b$, $c$ are distinct; (*ii*) $a = b \neq c$; (*iii*) $a = b = c$.
Printed answer:
- $z = 1 - x - y$ if $k=a$, inconsistent if $k\ne a$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page102

Exercise Page102, problem 1, p. 102
$\begin{System}[\,]{2} 8x &-{}& y &= 34, \\ x &+{}& 8y &= 53. \end{System}$
Printed answer:
- $x = 5$, $y = 6$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 5, y: 6}
Exercise Page102, problem 2, p. 102
$\begin{System}[\,]{2} 3x &+{}& 4y &= 10, \\ 4x &+{}& y &= 9. \end{System}$
Printed answer:
- $x = 2$, $y = 1$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: 2, y: 1}
Exercise Page102, problem 3, p. 102
$\begin{System}[\,]{2} ax &+{}& by &= a^2, \\ bx &-{}& ay &= ab. \end{System}$
Printed answer:
- $x = a$, $y = 0$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: a, y: 0}

Exercise Page104

The data holds no problems for this exercise yet.

Exercise Page106

Exercise Page106, problem 1, p. 106
Find the six terms involving $a_2$ in the determinant $(7)$.
Printed answer:
- $-a_2b_1c_3d_4 + a_2b_1c_4d_3 + a_2b_3c_1d_4 - a_2b_3c_4d_1 - a_2b_4c_1d_3 + a_2b_4c_3d_1$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page106, problem 2, p. 106
What are the signs of $a_3b_5c_2d_1e_4$, $a_5b_4c_3d_2e_1$ in a determinant of order five?
Printed answer:
- $+$, $+$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page106, problem 3, p. 106
Show that the arrangement $4, 1, 3, 2$ may be obtained from $1, 2, 3, 4$ by use of the two successive interchanges $(1, 4)$, $(1, 2)$, and also by use of the four successive interchanges $(1, 4)$, $(1, 3)$, $(1, 2)$, $(2, 3)$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page106, problem 4, p. 106
Write out the six terms of $(8)$ for $n = 3$, rearrange the factors of each term so that the new first subscripts shall be in the order $1, 2, 3$, and verify that the resulting six terms are those of the determinant $D'$ in §85 for $n = 3$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page108

The data holds no problems for this exercise yet.

Exercise Page112

Exercise Page112, problem 1, p. 112
$\ds \begin{vmatrix} 3a & 3b & 3c \\ 5a & 5b & 5c \\ d & e & f \end{vmatrix} = 0$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page112, problem 2, p. 112
$\ds \begin{vmatrix} 2r & l & 3r \\ 2s & m & 3s \\ 2t & n & 3t \end{vmatrix} = 0$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page112, problem 3, p. 112
$\ds \begin{vmatrix} 2 & 7 & 3 \\ 5 & 9 & 8 \\ 0 & 3 & 0 \end{vmatrix}$.
Printed answer:
- $-3$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes -3
On the STU-32 (STU, rpn):
```
2 ENTER 8 × 3 ENTER 5 × − 3 × +/−
```
Calculator: -3E+0; the book prints -3. Run on the calculator core at firmware 628c96c.
Exercise Page112, problem 4, p. 112
$\ds \begin{vmatrix} 5 & 7 & 0 \\ 6 & 8 & 0 \\ 3 & 9 & 4 \end{vmatrix}$.
Printed answer:
- $-8$.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes -8
On the STU-32 (STU, rpn):
```
5 ENTER 8 ×
```
```
7 ENTER 6 ×
```
```
−
```
```
4 ×
```
Calculator: -8E+0; the book prints -8. Run on the calculator core at firmware 628c96c.
Exercise Page112, problem 5, p. 112
$\ds \begin{vmatrix} a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ a^4 & b^4 & c^4 & d^4 \end{vmatrix} = abcd(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page113

The data holds no problems for this exercise yet.

Exercise Page120

Exercise Page120, problem 1, p. 120
$\begin{System}{3} x &+{}& y &+{}& 3z &= 0,\\ x &+{}& 2y &+{}& 2z &= 0,\\ x &+{}& 5y &-{}& z &= 0. \end{System}$
Printed answer:
- $r = 2$, $x:y:z = -4:1:1$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page120, problem 2, p. 120
$\begin{System}{3} 2x &-{}& y &+{}& 4z &= 0,\\ x &+{}& 3y &-{}& 2z &= 0,\\ x &-{}& 11y &+{}& 14z &= 0. \end{System}$
Printed answer:
- $r = 2$, $x:y:z = -10:8:7$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page120, problem 3, p. 120
$\begin{System}{3} x &-{}& 3y &+{}& 4z &= 0,\\ 4x &-{}& 12y &+{}& 16z &= 0,\\ 3x &-{}& 9y &+{}& 12z &= 0. \end{System}$
Printed answer:
- $r = 1$, two unknowns arbitrary.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page120, problem 4, p. 120
$\begin{System}{4} 6x &+{}& 4y &+{}& 3z &-{}& 84w &= 0,\\ x &+{}& 2y &+{}& 3z &-{}& 48w &= 0,\\ x &-{}& 2y &+{}& z &-{}& 12w &= 0,\\ 4x &+{}& 4y &-{}& z &-{}& 24w &= 0. \end{System}$
Printed answer:
- $r = 3$, $x:y:z:w = 6:3:12:1$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page120, problem 5, p. 120
$\begin{System}{4} 2x &+{}& 3y &-{}& 4z &+{}& 5w &= 0,\\ 3x &+{}& 5y &-{}& z &+{}& 2w &= 0,\\ 7x &+{}& 11y &-{}& 9z &+{}& 12w &= 0,\\ 3x &+{}& 4y &-{}& 11z &+{}& 13w &= 0. \end{System}$
Printed answer:
- $r = 2$, $z = -\frac{11}{3} x - \frac{19}{3} y$, $w = -\frac{10}{3} x - \frac{17}{3} y$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page121

Exercise Page121, problem 1, p. 121
$\begin{System}{3} 2x &+{}& y &+{}& 3z &= 1,\\ 4x &+{}& 2y &-{}& z &= -3,\\ 2x &+{}& y &-{}& 4z &= -4,\\ 10x &+{}& 5y &-{}& 6z &= -10. \end{System}$
Printed answer:
- Ranks of $A$ and $B$ are $2$; $y = -8/7 - 2x, z = 5/7$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page121, problem 2, p. 121
$\begin{System}{3} 2x &-{}& y &+{}& 3z &= 2,\\ x &+{}& 7y &+{}& z &= 1,\\ 3x &+{}& 5y &-{}& 5z &= a,\\ 4x &-{}& 3y &+{}& 2z &= 1. \end{System}$
Printed answer:
- Consistent only when $a = -225/61$ and then $x = -\dfrac{5}{61}$, $y = \dfrac{3}{61}$, $z = \dfrac{45}{61}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {"a": "-225/61", "x": "-5/61", "y": "3/61", "z": "45/61"}
Exercise Page121, problem 3, p. 121
$\begin{System}{3} 4x &-{}& y &+{}& z &= 5,\\ 2x &-{}& 3y &+{}& 5z &= 1,\\ x &+{}& y &-{}& 2z &= 2,\\ 5x & & &-{}& z &= 2. \end{System}$
Printed answer:
- Rank of $A$ is $2$, rank of $B$ is $3$, inconsistent.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page121, problem 4, p. 121
$\begin{System}{2} 4x &-{}& 5y &= 2,\\ 2x &+{}& 3y &= 12,\\ 10x &-{}& 7y &= 16. \end{System}$
Printed answer:
- $A$ and $B$ of rank $2$, $x = 3$, $y = 2$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {"x": 3, "y": 2}
Exercise Page121, problem 5, p. 121
Prove the Corollary by multiplying the known terms by $x_{n+1}=1$ and applying §97 with $n$ replaced by $n+1$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page121, problem 6, p. 121
Prove that if the matrix of the coefficients of any system of linear homogeneous % equations in $n$ unknowns is of rank $r$, the values of certain $n-r$ of the unknowns may be %% -----File: 128.png---Folio 122------- assigned at pleasure and the others will then be uniquely determined and satisfy all of the equations.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page124

The data holds no problems for this exercise yet.

Exercise Page125

The data holds no problems for this exercise yet.

Exercise Page126

Exercise Page126, problem 1, p. 126
Solve System3 ax &+& by &+& cz &= k, a^2x &+& b^2y &+& c^2z &= k^2, a^4x &+& b^4y &+& c^4z &= k^4 System by determinants for $x$, treating all cases.
Printed answer:
- $x = \dfrac{k(b-k)(c-k)(k+b+c)}{a(b-a)(c-a)(a+b+c)}$, if $a$, $b$, $c$ are distinct and not zero and their sum $\neq 0$. If $a = b \neq c$, $ac \ne 0$, equations are inconsistent unless $k = 0$, $a$, $c$, or $-a-c$, and then $y = \dfrac{k(c-k)}{a(c-a)} - x$, $z = \dfrac{k(k-a)}{c(c-a)}$, $x$ arbitrary.
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem k*(b-k)*(c-k)*(k+b+c)/(a*(b-a)*(c-a)*(a+b+c))
Exercise Page126, problem 10, p. 126
Prove that the cubic equation % D(x) vmatrix a-x & b & c b & f-x & g c & g & h-x vmatrix = 0 has only real roots. Hints: gather* D(x) · D(-x) = |arraylll a^2+b^2+c^2-x^2 & ab+bf+cg & ac+bg+ch ab+bf+cg & b^2+f^2+g^2-x^2 & bc+fg+gh ac+bg+ch & bc+fg+gh & c^2+g^2+h^2-x^2 array| % = -x^6+x^4(a^2+f^2+h^2+2b^2+2c^2+2g^2) - x^2(D_1+D_2+D_3)+ D^2(0), gather* where $D_3$ denotes the first determinant in Ex. 9 with all accents removed and with $e = b$, while $D_1$ and $D_2$ are analogous minors of elements in the main diagonal of the present determinant of order $3$ with $x = 0$. Hence the coefficient of $-x^2$ is a sum of squares. Since the function of degree $6$ is not zero for a negative value of $x^2$, $D(x)=0$ has no purely imaginary root. If it had an imaginary root $r+si$, then $D(x+r)=0$ would have a purely imaginary root $si$. But $D(x+r)$ is of the form $D(x)$ with $a$, $f$, $h$ replaced by $a-r$, $f-r$, $h-r$. Hence $D(x)=0$ has only real roots. The method is applicable to such determinants of order $n$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page126, problem 11, p. 126
If $a_1, \dotsc, a_n$ are distinct, solve the system of equations x_1k_i-a_1 + x_2k_i-a_2 + + x_nk_i - a_n = 1 (i=1, , n). Hint: Regard $k_1, \dotsc, k_n$ as the roots of an equation of degree $n$ in $k$ formed from the typical one above by substituting $k$ for $k_i$ and clearing of fractions; write $k = a_j-t$, and consider the product of the roots of $t^n + \dotsb = 0$. Hence find $x_j$.
Printed answer:
- $\ds x_j = (k_1-a_j)\dotsm(k_n-a_j) \div \prod\limits^n_{\substack{s=1 \\ s\neq j}} (a_s-a_j)$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page126, problem 12, p. 126
Solve the equation vmatrix a+x & x & x x & b+x & x x & x & c+x vmatrix = 0.
Printed answer:
- $x(ab + ac + bc) = -abc$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes -a*b*c/(a*b + a*c + b*c)
Exercise Page126, problem 2, p. 126
In three linear homogeneous equations in four unknowns, prove that the values of the unknowns are proportional to four determinants of order $3$ formed from the coefficients.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page126, problem 3, p. 126
$\ds \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$.
Printed answer:
- $(a-b)(b-c)(c-a)$.
verified: the printed answer passed a computed check
How it was checked
- factor: passes (a-b)*(b-c)*(c-a)
Exercise Page126, problem 4, p. 126
$\ds \begin{vmatrix} x & x^2 & yz \\ y & y^2 & xz \\ z & z^2 & xy \end{vmatrix} = \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix}$.
Printed answer:
- $(x-y)(y-z)(z-x)(xy + yz + zx)$.
verified: the printed answer passed a computed check
How it was checked
- factor: passes (x-y)*(y-z)*(z-x)*(x*y+y*z+z*x)
Exercise Page126, problem 5, p. 126
vmatrix a & b & c c & a & b b & c & a vmatrix = (a+b+c)(a+b+c^2)(a+b^2+c), where $\omega$ is an imaginary cube root of unity.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- factor: no printed answer to check
Exercise Page126, problem 6, p. 126
$\ds \begin{vmatrix} a & b & c & d \\ b & a & d & c \\ c & d & a & b \\ d & c & b & a \end{vmatrix}$.
Printed answer:
- $(a+b+c+d)(a+b-c-d)(a-b-c+d)(a-b+c-d)$.
verified: the printed answer passed a computed check
How it was checked
- factor: passes (a+b+c+d)*(a+b-c-d)*(a-b-c+d)*(a-b+c-d)
Exercise Page126, problem 7, p. 126
$\ds \begin{vmatrix} a & b & c & d \\ d & a & b & c \\ c & d & a & b \\ b & c & d & a \end{vmatrix}$.
Printed answer:
- $(a+b+c+d)(a-b+c-d)(a+bi-c-di)(a-bi-c+di)$.
unverified: no computed check settled this one (yet)
How it was checked
- factor: the printed answer does not match the problem (a+b+c+d)*(a-b+c-d)*(a+b*I-c-d*I)*(a-b*I-c+d*I)
Exercise Page126, problem 8, p. 126
If the points $(x_1, y_1), \dotsc, (x_4, y_4)$ lie on a circle, prove that |arraycccc x_1^2 + y_1^2 & x_1 & y_1 & 1 [2]4 x_4^2 + y_4^2 & x_4 & y_4 & 1 array| = 0.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page126, problem 9, p. 126
Prove that gather* vmatrix aa’ + bb’ + cc’ & ea’ + fb’ + gc’ ae’ + bf’ + cg’ & ee’ + ff’ + gg’ vmatrix % = vmatrix a & b e & f vmatrix · vmatrix a’ & b’ e’ & f’ vmatrix + vmatrix a & c e & g vmatrix · vmatrix a’ & c’ e’ & g’ vmatrix + vmatrix b & c f & g vmatrix · vmatrix b’ & c’ f’ & g’ vmatrix. gather*
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles