Chapter 4: DeterminantsContent check — Linear Algebra, Chapter 4: Determinants

$4$

$3$

$-12$

$6$

$21$

$27$

For the first, apply the formula in this section, note that any term with a $d$⁠, $g$⁠, or $h$ is zero, and simplify. Lower-triangular matrices work the same way.

Nonsingular, the determinant is $-1$⁠.

Nonsingular, the determinant is $-1$⁠.

Singular, the determinant is $0$⁠.

Nonsingular, the determinant is $3$⁠.

Singular, the determinant is $0$⁠.

Singular, the determinant is $0$⁠.

$\det (B)=\det (A)$ via $-2{\rho }_1+{\rho }_2$

$\det (B)=-\det (A)$ via ${\rho }_2\leftrightarrow {\rho }_3$

$\det (B)=(1/2)\cdot \det (A)$ via $(1/2){\rho }_2$

Gauss’s Method does this.

$$\left(\begin{array}{cccc}1& 2& 0& 2\\ 2& 4& 1& 0\\ 0& 0& -1& 3\\ 3& -1& 1& 4\end{array}\right)\underset{\begin{array}{c}-3{\rho }_1+{\rho }_4\end{array}}{\overset{-2{\rho }_1+{\rho }_2}{⟶}}\stackrel{{\rho }_2\leftrightarrow {\rho }_4}{⟶}\stackrel{{\rho }_3+{\rho }_4}{⟶}\left(\begin{array}{cccc}1& 2& 0& 2\\ 0& -7& 1& -2\\ 0& 0& -1& 3\\ 0& 0& 0& -1\end{array}\right)$$

The echelon form matrix has a product down the diagonal of $1\cdot (-7)\cdot (-1)\cdot (-1)=-7$⁠. In the course of Gauss’s Method no rows got rescaled but there was a row swap, so to get the determinant we change the sign, giving $+7$⁠.

Using the formula for the determinant of a $3\times 3$ matrix we expand the left side

$$1\cdot b\cdot c^2+1\cdot c\cdot a^2+1\cdot a\cdot b^2-b^2\cdot c\cdot 1-c^2\cdot a\cdot 1-a^2\cdot b\cdot 1$$

and by distributing we expand the right side.

$$(bc-ba-ac+a^2)\cdot (c-b)=c^{2}b-b^{2}c-bac+b^{2}a-a{c}^2+acb+a^{2}c-a^{2}b$$

Now we can just check that the two are equal. (Remark. This is the $3\times 3$ case of Vandermonde’s determinant which arises in applications).

This equation

$$0=\det (\left(\begin{array}{cc}12-x& 4\\ 8& 8-x\end{array}\right))=64-20x+x^2=(x-16)(x-4)$$

has roots $x=16$ and $x=4$⁠.

We first reduce the matrix to echelon form. To begin, assume that $a\ne 0$ and that $ae-bd\ne 0$⁠.

$$\begin{array}{c}\stackrel{(1/a){\rho }_1}{⟶}\left(\begin{array}{ccc}1& b/a& c/a\\ d& e& f\\ g& h& i\end{array}\right)\underset{\begin{array}{c}-g{\rho }_1+{\rho }_3\end{array}}{\overset{-d{\rho }_1+{\rho }_2}{⟶}}\left(\begin{array}{ccc}1& b/a& c/a\\ 0& (ae-bd)/a& (af-cd)/a\\ 0& (ah-bg)/a& (ai-cg)/a\end{array}\right)\\ \stackrel{(a/(ae-bd)){\rho }_2}{⟶}\left(\begin{array}{ccc}1& b/a& c/a\\ 0& 1& (af-cd)/(ae-bd)\\ 0& (ah-bg)/a& (ai-cg)/a\end{array}\right)\end{array}$$

This step finishes the calculation.

$$\stackrel{((ah-bg)/a){\rho }_2+{\rho }_3}{⟶}\left(\begin{array}{ccc}1& b/a& c/a\\ 0& 1& (af-cd)/(ae-bd)\\ 0& 0& (aei+bgf+cdh-hfa-idb-gec)/(ae-bd)\end{array}\right)$$

Now assuming that $a\ne 0$ and $ae-bd\ne 0$⁠, the original matrix is nonsingular if and only if the $\mathrm{3,3}$ entry above is nonzero. That is, under the assumptions, the original matrix is nonsingular if and only if $aei+$ $bgf+$ $cdh-$ $hfa-$ $idb-$ $gec\ne 0$⁠, as required.

We finish by running down what happens if the assumptions that were taken for convenience in the prior paragraph do not hold. First, if $a\ne 0$ but $ae-bd=0$ then we can swap

$$\left(\begin{array}{ccc}1& b/a& c/a\\ 0& 0& (af-cd)/a\\ 0& (ah-bg)/a& (ai-cg)/a\end{array}\right)\stackrel{{\rho }_2\leftrightarrow {\rho }_3}{⟶}\left(\begin{array}{ccc}1& b/a& c/a\\ 0& (ah-bg)/a& (ai-cg)/a\\ 0& 0& (af-cd)/a\end{array}\right)$$

and conclude that the matrix is nonsingular if and only if either $ah-bg=0$ or $af-cd=0$⁠. The condition ‘⁠$ah-bg=0$ or $af-cd=0$⁠’ is equivalent to the condition ‘⁠$(ah-bg)(af-cd)=0$⁠’. Multiplying out and using the case assumption that $ae-bd=0$ to substitute $ae$ for $bd$ gives this.

$$\begin{array}{c}0=ahaf-ahcd-bgaf+bgcd=ahaf-ahcd-bgaf+aegc\\ =a(haf-hcd-bgf+egc)\end{array}$$

Since $a\ne 0$⁠, we have that the matrix is nonsingular if and only if $haf-hcd-bgf+egc=0$⁠. Therefore, in this $a\ne 0$ and $ae-bd=0$ case, the matrix is nonsingular when $haf-$ $hcd-$ $bgf+$ $egc-$ $i(ae-bd)=$ $0$⁠.

The remaining cases are routine. Do the $a=0$ but $d\ne 0$ case and the $a=0$ and $d=0$ but $g\ne 0$ case by first swapping rows and then going on as above. The $a=0$⁠, $d=0$⁠, and $g=0$ case is easy — that matrix is singular since the columns form a linearly dependent set, and the determinant comes out to be zero.

Figuring the determinant and doing some algebra gives this.

$$\begin{array}{cc}\hfill 0& =y_{1}x+x_{2}y+x_1{y}_2-y_{2}x-x_{1}y-x_2{y}_1\hfill \\ \hfill (x_2-x_1)\cdot y& =(y_2-y_1)\cdot x+x_2{y}_1-x_1{y}_2\hfill \\ \hfill y& =\frac{y_2-y_1}{x_2-x_1}\cdot x+\frac{x_2{y}_1-x_1{y}_2}{x_2-x_1}\hfill \end{array}$$

Note that this is the equation of a line (in particular, in contains the familiar expression for the slope), and note that $(x_1,y_1)$ and $(x_2,y_2)$ satisfy it.

The comparison with the formula given in the preamble to this section is easy.

While it holds for $2\times 2$ matrices

$$\begin{array}{cc}\hfill \left(\begin{array}{ccc}{h}_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,1}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,1}}\end{array}\right)& =\begin{array}{c}{h}_{\mathrm{1,1}}{h}_{\mathrm{2,2}}+h_{\mathrm{1,2}}{h}_{\mathrm{2,1}}\hfill \\ -h_{\mathrm{2,1}}{h}_{\mathrm{1,2}}-h_{\mathrm{2,2}}{h}_{\mathrm{1,1}}\hfill \end{array}\hfill \\ \hfill & =h_{\mathrm{1,1}}{h}_{\mathrm{2,2}}-h_{\mathrm{1,2}}{h}_{\mathrm{2,1}}\hfill \end{array}$$

it does not hold for $4\times 4$ matrices. An example is that this matrix is singular because the second and third rows are equal

$$\left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 1& 1& 0\\ -1& 0& 0& 1\end{array}\right)$$

but following the scheme of the mnemonic does not give zero.

$$\left(\begin{array}{ccccccc}\hfill 1& \hfill 0& \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 1& \hfill 1\\ \hfill 0& \hfill 1& \hfill 1& \hfill 0& \hfill 0& \hfill 1& \hfill 1\\ \hfill -1& \hfill 0& \hfill 0& \hfill 1& \hfill -1& \hfill 0& \hfill 0\end{array}\right)=\begin{array}{c}1+0+0+0\hfill \\ -(-1)-0-0-0\hfill \end{array}$$

The determinant is $(x_2{y}_3-x_3{y}_2){\overrightarrow{e}}_1+$ $(x_3{y}_1-x_1{y}_3){\overrightarrow{e}}_2+$ $(x_1{y}_2-x_2{y}_1){\overrightarrow{e}}_3$⁠. To check perpendicularity, we check that the dot product with the first vector is zero

$$\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)\bullet \left(\begin{array}{c}{x}_2{y}_3-x_3{y}_2\\ x_3{y}_1-x_1{y}_3\\ x_1{y}_2-x_2{y}_1\end{array}\right)=x_1{x}_2{y}_3-x_1{x}_3{y}_2+x_2{x}_3{y}_1-x_1{x}_2{y}_3+x_1{x}_3{y}_2-x_2{x}_3{y}_1=0$$

and the dot product with the second vector is also zero.

$$\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right)\bullet \left(\begin{array}{c}{x}_2{y}_3-x_3{y}_2\\ x_3{y}_1-x_1{y}_3\\ x_1{y}_2-x_2{y}_1\end{array}\right)=x_2{y}_1{y}_3-x_3{y}_1{y}_2+x_3{y}_1{y}_2-x_1{y}_2{y}_3+x_1{y}_2{y}_3-x_2{y}_1{y}_3=0$$

Plug and chug: the determinant of the product is this

$$\begin{array}{cc}\hfill \det (\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}w& x\\ y& z\end{array}\right))& =\det (\left(\begin{array}{cc}aw+by& ax+bz\\ cw+dy& cx+dz\end{array}\right))\hfill \\ \hfill & =\begin{array}{c}acwx+adwz+bcxy+bdyz\hfill \\ -acwx-bcwz-adxy-bdyz\hfill \end{array}\hfill \end{array}$$

while the product of the determinants is this.

$$\det (\left(\begin{array}{cc}a& b\\ c& d\end{array}\right))\cdot \det (\left(\begin{array}{cc}w& x\\ y& z\end{array}\right))=(ad-bc)\cdot (wz-xy)$$

Verification that they are equal is easy.

Use the prior item.

That similar matrices have the same determinant is immediate from the above two: $\det (PT{P}^{-1})=$ $\det (P)\cdot \det (T)\cdot \det (P^{-1})$⁠.

One way is to count these areas

by taking the area of the entire rectangle and subtracting the area of $A$ the upper-left rectangle, $B$ the upper-middle triangle, $D$ the upper-right triangle, $C$ the lower-left triangle, $E$ the lower-middle triangle, and $F$ the lower-right rectangle $(x_1+x_2)(y_1+y_2)-$ $x_2{y}_1-$ $(1/2)x_1{y}_1-$ $(1/2)x_2{y}_2-$ $(1/2)x_2{y}_2-$ $(1/2)x_1{y}_1-$ $x_2{y}_1$⁠. Simplification gives the determinant formula.

This determinant is the negative of the one above; the formula distinguishes whether the second column is counterclockwise from the first.

The computation for $2\times 2$ matrices, using the formula quoted in the preamble, is easy. It does also hold for $3\times 3$ matrices; the computation is routine.

No. We illustrate with the $2\times 2$ determinant. Recall that constants come out one row at a time.

$$\det (\left(\begin{array}{cc}2& 4\\ 2& 6\end{array}\right))=2\cdot \det (\left(\begin{array}{cc}1& 2\\ 2& 6\end{array}\right))=2\cdot 2\cdot \det (\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right))$$

This contradicts linearity (here we didn’t need $S$⁠, i.e., we can take $S$ to be the matrix of zeros).

Bring out the $c$⁠’s one row at a time.

There are no real numbers $\theta$ that make the matrix singular because the determinant of the matrix ${\cos }^2\theta +{\sin }^2\theta$ is never $0$⁠, it equals $1$ for all $\theta$⁠. Geometrically, with respect to the standard basis, this matrix represents a rotation of the plane through an angle of $\theta$⁠. Each such map is one-to-one — for one thing, it is invertible.

This is how the answer was given in the cited source. Let $P$ be the sum of the three positive terms of the determinant and $-N$ the sum of the three negative terms. The maximum value of $P$ is

$$9\cdot 8\cdot 7+6\cdot 5\cdot 4+3\cdot 2\cdot 1=630.$$

The minimum value of $N$ consistent with $P$ is

$$9\cdot 6\cdot 1+8\cdot 5\cdot 2+7\cdot 4\cdot 3=218.$$

Any change in $P$ would result in lowering that sum by more than $4$⁠. Therefore $412$ the maximum value for the determinant and one form for the determinant is

$$\left|\begin{array}{ccc}9& 4& 2\\ 3& 8& 6\\ 5& 1& 7\end{array}\right|.$$

Do $2{\rho }_1+{\rho }_2$ to get echelon form, and then multiply down the diagonal. The determinant is $0$⁠.

Swapping the second and third rows brings the system to echelon form (and changes the sign of the determinant). Multiplying down the diagonal gives $12$⁠, so the determinant of the given matrix is $-12$⁠.

$\left|\begin{array}{ccc}3& 1& 2\\ 3& 1& 0\\ 0& 1& 4\end{array}\right|=$ $\left|\begin{array}{ccc}3& 1& 2\\ 0& 0& -2\\ 0& 1& 4\end{array}\right|=$ $-$ $\left|\begin{array}{ccc}3& 1& 2\\ 0& 1& 4\\ 0& 0& -2\end{array}\right|=$ $6$

$\left|\begin{array}{cccc}1& 0& 0& 1\\ 2& 1& 1& 0\\ -1& 0& 1& 0\\ 1& 1& 1& 0\end{array}\right|=$ $\left|\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& -2\\ 0& 0& 1& 1\\ 0& 1& 1& -1\end{array}\right|=$ $\left|\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& -2\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}\right|=$ $1$

$\left|\begin{array}{cc}2& -1\\ -1& -1\end{array}\right|=\left|\begin{array}{cc}2& -1\\ 0& -3/2\end{array}\right|=-3$⁠;

$\left|\begin{array}{ccc}1& 1& 0\\ 3& 0& 2\\ 5& 2& 2\end{array}\right|=$ $\left|\begin{array}{ccc}1& 1& 0\\ 0& -3& 2\\ 0& -3& 2\end{array}\right|=$ $\left|\begin{array}{ccc}1& 1& 0\\ 0& -3& 2\\ 0& 0& 0\end{array}\right|=$ $0$

When is the determinant not zero?

$$\left|\begin{array}{cccc}1& 0& 1& -1\\ 0& 1& -2& 0\\ 1& 0& k& 0\\ 0& 0& 1& -1\end{array}\right|=\left|\begin{array}{cccc}1& 0& 1& -1\\ 0& 1& -2& 0\\ 0& 0& k-1& 1\\ 0& 0& 1& -1\end{array}\right|$$

Obviously, $k=1$ gives nonsingularity and hence a nonzero determinant. If $k\ne 1$ then we get echelon form with a $(-1/k-1){\rho }_3+{\rho }_4$ combination.

$$=\left|\begin{array}{cccc}1& 0& 1& -1\\ 0& 1& -2& 0\\ 0& 0& k-1& 1\\ 0& 0& 0& -1-(1/k-1)\end{array}\right|$$

Multiplying down the diagonal gives $(k-1)(-1-(1/k-1))=$ $-$ $(k-1)-$ $1=$ $-$ $k$⁠. Thus the matrix has a nonzero determinant, and so the system has a unique solution, if and only if $k\ne 0$⁠.

Condition (2) of the definition of determinants applies via the swap ${\rho }_1\leftrightarrow {\rho }_3$⁠.

$$\left|\begin{array}{ccc}{h}_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,3}}\end{array}\right|=-\left|\begin{array}{ccc}{h}_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\end{array}\right|$$

Condition (3) applies.

$$\begin{array}{c}\left|\begin{array}{ccc}-h_{\mathrm{1,1}}& -h_{\mathrm{1,2}}& -h_{\mathrm{1,3}}\\ -2h_{\mathrm{2,1}}& -2h_{\mathrm{2,2}}& -2h_{\mathrm{2,3}}\\ -3h_{\mathrm{3,1}}& -3h_{\mathrm{3,2}}& -3h_{\mathrm{3,3}}\end{array}\right|=(-1)\cdot (-2)\cdot (-3)\cdot \left|\begin{array}{ccc}{h}_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\end{array}\right|\\ =(-6)\cdot \left|\begin{array}{ccc}{h}_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\end{array}\right|\end{array}$$

$$\begin{array}{c}\left|\begin{array}{ccc}{h}_{\mathrm{1,1}}+h_{\mathrm{3,1}}& h_{\mathrm{1,2}}+h_{\mathrm{3,2}}& h_{\mathrm{1,3}}+h_{\mathrm{3,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ 5h_{\mathrm{3,1}}& 5h_{\mathrm{3,2}}& 5h_{\mathrm{3,3}}\end{array}\right|\\ \begin{array}{cc}\hfill & =5\cdot \left|\begin{array}{ccc}{h}_{\mathrm{1,1}}+h_{\mathrm{3,1}}& h_{\mathrm{1,2}}+h_{\mathrm{3,2}}& h_{\mathrm{1,3}}+h_{\mathrm{3,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\end{array}\right|\hfill \\ \hfill & =5\cdot \left|\begin{array}{ccc}{h}_{\mathrm{1,1}}& h_{\mathrm{1,2}}& h_{\mathrm{1,3}}\\ h_{\mathrm{2,1}}& h_{\mathrm{2,2}}& h_{\mathrm{2,3}}\\ h_{\mathrm{3,1}}& h_{\mathrm{3,2}}& h_{\mathrm{3,3}}\end{array}\right|\hfill \end{array}\end{array}$$

A diagonal matrix is in echelon form, so the determinant is the product down the diagonal.

It is the trivial subspace.

Adding the second row to the first gives a matrix whose first row is $x+y+z$ times its third row.

$\left(\begin{array}{c}1\end{array}\right)$⁠, $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$⁠, $\left(\begin{array}{ccc}1& -1& 1\\ -1& 1& -1\\ 1& -1& 1\end{array}\right)$

The determinant in the $1\times 1$ case is $1$⁠. In every other case the second row is the negative of the first, and so matrix is singular and the determinant is zero.

$\left(\begin{array}{c}2\end{array}\right)$⁠, $\left(\begin{array}{cc}2& 3\\ 3& 4\end{array}\right)$⁠, $\left(\begin{array}{ccc}2& 3& 4\\ 3& 4& 5\\ 4& 5& 6\end{array}\right)$

The $1\times 1$ and $2\times 2$ cases yield these.

$$\left|\begin{array}{c}2\end{array}\right|=2\left|\begin{array}{cc}2& 3\\ 3& 4\end{array}\right|=-1$$

And $n\times n$ matrices with $n\ge 3$ are singular, e.g.,

$$\left|\begin{array}{ccc}2& 3& 4\\ 3& 4& 5\\ 4& 5& 6\end{array}\right|=0$$

because twice the second row minus the first row equals the third row. Checking this is routine.

This one

$$A=B=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$$

is easy to check.

$$|A+B|=\left|\begin{array}{cc}2& 4\\ 6& 8\end{array}\right|=-8\left|A\right|+\left|B\right|=-2-2=-4$$

By the way, this also gives an example where scalar multiplication is not preserved $|2\cdot A|\ne 2\cdot \left|A\right|$⁠.

No, we cannot replace it. Remark 2.2 shows that the four conditions after the replacement would conflict — no function satisfies all four.

A upper-triangular matrix is in echelon form.

A lower-triangular matrix is either singular or nonsingular. If it is singular then it has a zero on its diagonal and so its determinant (namely, zero) is indeed the product down its diagonal. If it is nonsingular then it has no zeroes on its diagonal, and we can reduce it by Gauss’s Method to echelon form without changing the diagonal.

The properties in the definition of determinant show that $\left|M_i(k)\right|=k$⁠, $\left|P_{i,j}\right|=-1$⁠, and $\left|C_{i,j}(k)\right|=1$⁠.

The three cases are easy to check by recalling the action of left multiplication by each type of matrix.

If $TS$ is invertible $(TS)M=I$ then the associative property of matrix multiplication $T(SM)=I$ shows that $T$ is invertible. So if $T$ is not invertible then neither is $TS$⁠.

If $T$ is singular then apply the prior answer: $\left|TS\right|=0$ and $\left|T\right|\cdot \left|S\right|=0\cdot \left|S\right|=0$⁠. If $T$ is not singular then we can write it as a product of elementary matrices $\left|TS\right|=$ $|E_r\cdots E_{1}S|=$ $\left|E_r\right|\cdots \left|E_1\right|\cdot \left|S\right|=$ $|E_r\cdots E_1|\left|S\right|=$ $\left|T\right|\left|S\right|$⁠.

$1=\left|I\right|=|T\cdot T^{-1}|=\left|T\right|\left|T^{-1}\right|$

We must show that if

$$T\stackrel{k{\rho }_i+{\rho }_j}{⟶}\hat{T}$$

then $d(T)=$ $\left|TS\right|/\left|S\right|=$ $\left|\hat{T}S\right|/\left|S\right|=$ $d(\hat{T})$⁠. We will be done if we show that combining rows first and then multiplying to get $\hat{T}S$ gives the same result as multiplying first to get $TS$ and then combining (because the determinant $\left|TS\right|$ is unaffected by the combination so we’ll then have $\left|\hat{T}S\right|=\left|TS\right|$⁠, and hence $d(\hat{T})=d(T)$⁠). That argument runs: after adding $k$ times row $i$ of $TS$ to row $j$ of $TS$⁠, the $j,p$ entry is $(k{t}_{i,1}+t_{j,1})s_{1,p}+$ $\cdots +$ $(k{t}_{i,r}+t_{j,r})s_{r,p}$⁠, which is the $j,p$ entry of $\hat{T}S$⁠.

We need only show that swapping $T\stackrel{{\rho }_i\leftrightarrow {\rho }_j}{⟶}\hat{T}$ and then multiplying to get $\hat{T}S$ gives the same result as multiplying $T$ by $S$ and then swapping (because, as the determinant $\left|TS\right|$ changes sign on the row swap, we’ll then have $\left|\hat{T}S\right|=-\left|TS\right|$⁠, and so $d(\hat{T})=-d(T)$⁠). That argument runs just like the prior one.

Not surprisingly by now, we need only show that multiplying a row by a scalar $T\stackrel{k{\rho }_i}{⟶}\hat{T}$ and then computing $\hat{T}S$ gives the same result as first computing $TS$ and then multiplying the row by $k$ (as the determinant $\left|TS\right|$ is rescaled by $k$ the multiplication, we’ll have $\left|\hat{T}S\right|=k\left|TS\right|$⁠, so $d(\hat{T})=kd(T)$⁠). The argument runs just as above.

Clear.

Because we’ve shown that $d(T)$ is a determinant and that determinant functions (if they exist) are unique, we have that so $\left|T\right|=d(T)=\left|TS\right|/\left|S\right|$⁠.

We will first argue that a rank $r$ matrix has a $r\times r$ submatrix with nonzero determinant. A rank $r$ matrix has a linearly independent set of $r$ rows. A matrix made from those rows will have row rank $r$ and thus has column rank $r$⁠. Conclusion: from those $r$ rows we can extract a linearly independent set of $r$ columns, and so the original matrix has a $r\times r$ submatrix of rank $r$⁠.

We finish by showing that if $r$ is the largest such integer then the rank of the matrix is $r$⁠. We need only show, by the maximality of $r$⁠, that if a matrix has a $k\times k$ submatrix of nonzero determinant then the rank of the matrix is at least $k$⁠. Consider such a $k\times k$ submatrix. Its rows are parts of the rows of the original matrix, clearly the set of whole rows is linearly independent. Thus the row rank of the original matrix is at least $k$⁠, and the row rank of a matrix equals its rank.

A matrix with only rational entries reduces with Gauss’s Method to an echelon form matrix using only rational arithmetic. Thus the entries on the diagonal must be rationals, and so the product down the diagonal is rational.

This is how the answer was given in the cited source. The value ${(1-a^4)}^3$ of the determinant is independent of the values $B$⁠, $C$⁠, $D$⁠. Hence operation (e) does not change the value of the determinant but merely changes its appearance. Thus the element of likeness in (a), (b), (c), (d), and (e) is only that the appearance of the principle entity is changed. The same element appears in (f) changing the name-label of a rose, (g) writing a decimal integer in the scale of $12$⁠, (h) gilding the lily, (i) whitewashing a politician, and (j) granting an honorary degree.