Ch.13 Determinants

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Properties

Recall an n×nn\times n matrix TT is nonsingular iff

We work towards a formula to determine if a matrix is nonsingular

For a 1×11\times1 matrix

(a)(a) is trivially nonsingular iff a≠0a\ne0

For a 2×22\times2 matrix

(abcd)\begin{pmatrix}a&b\\c&d\end{pmatrix} is nonsingular iff ad−bc≠0ad-bc\ne0

For a 3×33\times3 matrix

(abcdefghi)\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix} is nonsingular iff aei+bfg+cdh−hfa−idb−gec≠0aei+bfg+cdh-hfa-idb-gec\ne0

We call the family of formulas aa, ad−bcad-bc, etc for each n×nn\times n matrix.
The determinant function detn×n:Mn×n→R\text{det}_{n\times n}:\mathcal{M}_{n\times n}\to\mathbb{R} is defined for each nn such that an n×nn\times n matrix TT is nonsingular iff detn×n(T)≠0\text{det}_{n\times n}(T)\ne0

Based on the first three determinants, we extrapolate conditions that the determinant function must satisfy:
if TT has rows ρ⃗1,...,ρ⃗n\vec{\rho}_1,...,\vec{\rho}_n

  1. det(ρ⃗1,...,kρ⃗i+ρ⃗j,...,ρ⃗n)=det(ρ⃗1,...,ρ⃗j,...,ρ⃗n)\text{det}(\vec{\rho}_1,...,k\vec{\rho}_i+\vec{\rho}_j,...,\vec{\rho}_n)=\text{det}(\vec{\rho}_1,...,\vec{\rho}_j,...,\vec{\rho}_n) for i≠ji\ne j
    (row combination operations don't change the determinant)
  2. det(ρ⃗1,...,ρ⃗j,...,ρ⃗i,...,ρ⃗n)=−det(ρ⃗1,...,ρ⃗i,...,ρ⃗j,...,ρ⃗n)\text{det}(\vec{\rho}_1,...,\vec{\rho}_j,...,\vec{\rho}_i,...,\vec{\rho}_n)=-\text{det}(\vec{\rho}_1,...,\vec{\rho}_i,...,\vec{\rho}_j,...,\vec{\rho}_n) for i≠ji\ne j
    (swapping rows makes the determinant negative)
  3. det(ρ⃗1,...,kρ⃗i,...,ρ⃗n)=k⋅det(ρ⃗1,...,ρ⃗i,...,ρ⃗n)\text{det}(\vec{\rho}_1,...,k\vec{\rho}_i,...,\vec{\rho}_n)=k\cdot\text{det}(\vec{\rho}_1,...,\vec{\rho}_i,...,\vec{\rho}_n) for any scalar kk (including k=0k=0)
    (multiplying a row by kk, multiplies the determinant by kk)
  4. det(I)=1\text{det}(I)=1 for identity matrix II

We often write ∣T∣|T| instead of det(T)\text{det}(T)

Small Note

(2) is redundant because
T→ρi+ρj →−ρj+ρi →ρi+ρj →−ρiT^T\xrightarrow{\rho_i+\rho_j}\space\xrightarrow{-\rho_j+\rho_i}\space\xrightarrow{\rho_i+\rho_j}\space\xrightarrow{-\rho_i}\hat{T}
This swaps the rows, and the first three operations don't change the determinant while the last negates it.

From above, we can derive these lemmas:

Proof

For the first, swap the two identical rows. By condition (2), the determinant is the opposite but the matrix is the same, so it must be 00.

For the second, multiply the zero row by 22. By condition (3), the determinant is doubled, but the matrix remains the same, so the determinant is the same. Thus it must be 00.

The third is by definition.

The fourth sentence has two cases: if it is singular, then it has a zero row, and contains a 00 in the diagonal. The determinant is 00 since it is singular, and equals the product down the diagonal.
If the echelon form matrix is nonsingular then none of the diagonal entries are 00. We can then use condition (3) to get ones in the diagonal:
∣t1,1t1,2⋯t1,n0t2,2t2,n⋮00tn,n∣=t1,1⋅t2,2⋯tn,n⋅∣1t1,2/t1,1⋯t1,n/t1,101t2,n/t2,2⋮001∣\left|\begin{matrix}t_{1,1}&t_{1,2}&\cdots&t_{1,n}\\0&t_{2,2}&&t_{2,n}\\\vdots\\0&0&&t_{n,n}\end{matrix}\right|=t_{1,1}\cdot t_{2,2}\cdots t_{n,n}\cdot\left|\begin{matrix}1&t_{1,2}/t_{1,1}&\cdots&t_{1,n}/t_{1,1}\\0&1&&t_{2,n}/t_{2,2}\\\vdots\\0&0&&1\end{matrix}\right|
Then, clearing out the columns uses condition (1), so
=t1,1⋅t2,2⋯tn,n⋅∣I∣=t1,1⋅t2,2⋯tn,n=t_{1,1}\cdot t_{2,2}\cdots t_{n,n}\cdot |I|=t_{1,1}\cdot t_{2,2}\cdots t_{n,n}
So, the determinant is the product down the diagonal in this case as well.

With these rules, we can find the determinant using Gauss's Method

Example 13.1

Using Gauss's Method, find the determinant of the matrix
(13−22043−15)\begin{pmatrix}1&3&-2\\2&0&4\\3&-1&5\end{pmatrix}

∣13−22043−15∣=∣13−20−6−80−10−11∣=∣13−20−6−800−7/3∣=1⋅(−6)⋅(−7/3)=14\begin{vmatrix}1&3&-2\\2&0&4\\3&-1&5\end{vmatrix}=\begin{vmatrix}1&3&-2\\0&-6&-8\\0&-10&-11\end{vmatrix}=\begin{vmatrix}1&3&-2\\0&-6&-8\\0&0&-7/3\end{vmatrix}=1\cdot(-6)\cdot(-7/3)=14

Example 13.2

Find the determinant of
(031120152)\begin{pmatrix}0&3&1\\1&2&0\\1&5&2\end{pmatrix}

Illustrate condition (2)
∣031120152∣=−∣120031152∣=−∣120031032∣=−∣120031001∣=−1⋅3⋅1=−3\begin{vmatrix}0&3&1\\1&2&0\\1&5&2\end{vmatrix}=-\begin{vmatrix}1&2&0\\0&3&1\\1&5&2\end{vmatrix}=-\begin{vmatrix}1&2&0\\0&3&1\\0&3&2\end{vmatrix}=-\begin{vmatrix}1&2&0\\0&3&1\\0&0&1\end{vmatrix}=-1\cdot3\cdot1=-3

For a 2×22\times2 determinant, notice how the terms are diagonals of the matrix.
In a 3×33\times3 matrix, the following mnemonics can be used

For larger matrices, use Gauss's Method

An upper triangular matrix is a square matrix with only 00s below the diagonal.
(t1,1t1,2⋯t1,n0t2,2⋯t2,n⋮⋮⋱⋮00⋯tn,n)\begin{pmatrix}t_{1,1}&t_{1,2}&\cdots&t_{1,n}\\0&t_{2,2}&\cdots&t_{2,n}\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&t_{n,n}\end{pmatrix}

The determinant of an upper triangular matrix TT is the product of the diagonals.

Proof

If the diagonal entries are all nonzero, the matrix is in echelon form, so the determinant is the product of the diagonals.
If ti,i=0t_{i,i}=0 for some i∈{1,...,n}i\in\{1,...,n\}, we must prove that ∣T∣=0|T|=0. This is true iff TT is singular, so it suffices to show that the columns of TT are linearly dependent. Consider the matrix formed by the first ii columns. It has the form
(t1,1t1,2⋯t1,i−1t1,i0t2,2⋯t2,i−1t2,i⋮⋮⋱⋮⋮00⋯ti−1,i−1ti−1,100⋯00⋮⋮⋱⋮⋮00⋯00)\begin{pmatrix}t_{1,1}&t_{1,2}&\cdots&t_{1,i-1}&t_{1,i}\\0&t_{2,2}&\cdots&t_{2,i-1}&t_{2,i}\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&t_{i-1,i-1}&t_{i-1,1}\\0&0&\cdots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\cdots&0&0\end{pmatrix}
Only the first i−1i-1 rows can be nonzero, therefore this matrix is singular. That implies the columns are linearly dependent, but the columns are a subset of the columns of TT, so the columns of TT must be linearly dependent.

Permutation Expansion

Existance and Uniqueness

The problem with using conditions to define a function is that we must verify that there is one and only one function that satisfies those conditions.
We do that by determining a well-defined formula.

First, prove its uniqueness:
For each nn, if there is an n×nn\times n determinant function than it is unique.

Proof

Suppose there existed two functions det1,det2:Mn×n→R\text{det}_1,\text{det}_2:\mathcal{M}_{n\times n}\to\mathbb{R} satisfying the four conditions. Given a square matrix MM, we can fix some way to reduce it to echelon form, keeping track of sign changes and scaler factors, then multiplying down the diagonal at the end, we can see that both functions must return the same result. Since they return the same output for every function, they are the same function.

Let VV be a vector space. A map f:Vn→Rf:V^n\to\mathbb{R} is multilinear if

  1. f(ρ⃗1,...,v⃗+w⃗,...,ρ⃗n)=f(ρ⃗1,...,v⃗,...,ρ⃗n)+f(ρ⃗1,...,+w⃗,...,ρ⃗n)f(\vec{\rho}_1,...,\vec{v}+\vec{w},...,\vec{\rho}_n)=f(\vec{\rho}_1,...,\vec{v},...,\vec{\rho}_n)+f(\vec{\rho}_1,...,+\vec{w},...,\vec{\rho}_n)
    (the function splits addition one input at a time)
  2. f(ρ⃗1,...,kv⃗,...,ρ⃗n)=k⋅f(ρ⃗1,...,v⃗,...,ρ⃗n)f(\vec{\rho}_1,...,k\vec{v},...,\vec{\rho}_n)=k\cdot f(\vec{\rho}_1,...,\vec{v},...,\vec{\rho}_n)
    (the function splits scalar multiples one input at a time)

Determinants are multilinear

Proof

The second of the two properties is simply condition (3) from above.
For the first condition, there are two conditions:
If the set of the other rows {ρ⃗1,...,ρ⃗i−1,ρ⃗i+1,...,ρ⃗n}\{\vec{\rho}_1,...,\vec{\rho}_{i-1},\vec{\rho}_{i+1},...,\vec{\rho}_n\} is linearly dependent, all three matrices are singular so we get the trivial 0=00=0.
Therefore, assume the set of other rows is linearly independent. Then we can add another vector to make a basis:
{ρ⃗1,...,ρ⃗i−1,β⃗,ρ⃗i+1,...,ρ⃗n}\{\vec{\rho}_1,...,\vec{\rho}_{i-1},\vec{\beta},\vec{\rho}_{i+1},...,\vec{\rho}_n\}
Then v⃗\vec{v} and w⃗\vec{w} can be expressed with respect to this basis and added:
v⃗=v1ρ⃗1+⋯+vi−1ρ⃗i−1+viβ⃗+vi+1ρ⃗i+1+⋯+vnρ⃗nw⃗=w1ρ⃗1+⋯+wi−1ρ⃗i−1+wiβ⃗+wi+1ρ⃗i+1+⋯+wnρ⃗nv⃗+w⃗=(v1+w1)ρ⃗1+⋯+(vi+wi)β⃗+⋯+(vn+wn)ρ⃗n\vec{v}=v_1\vec{\rho}_1+\cdots+v_{i-1}\vec{\rho}_{i-1}+v_i\vec{\beta}+v_{i+1}\vec{\rho}_{i+1}+\cdots+v_n\vec{\rho}_n\\\vec{w}=w_1\vec{\rho}_1+\cdots+w_{i-1}\vec{\rho}_{i-1}+w_i\vec{\beta}+w_{i+1}\vec{\rho}_{i+1}+\cdots+w_n\vec{\rho}_n\\\vec{v}+\vec{w}=(v_1+w_1)\vec{\rho}_1+\cdots+(v_i+w_i)\vec{\beta}+\cdots+(v_n+w_n)\vec{\rho}_n
Now substitute this into the left-hand side of property 1
det(ρ⃗1,...,(v1+w1)ρ⃗1+⋯+(vi+wi)β⃗+⋯+(vn+wn)ρ⃗n,...,ρ⃗n)\text{det}(\vec{\rho}_1,...,(v_1+w_1)\vec{\rho}_1+\cdots+(v_i+w_i)\vec{\beta}+\cdots+(v_n+w_n)\vec{\rho}_n,...,\vec{\rho}_n)
From condition (1), the determinant doesn't change if we add −(v⃗j+w⃗j)-(\vec{v}_j+\vec{w}_j) times ρ⃗j\vec{\rho}_j, so doing that we can apply condition (3) to get
det(ρ⃗1,...,v⃗+w⃗,...,ρ⃗n)=det(ρ⃗1,...,(vi+wi)β⃗,...,ρ⃗n)=(v⃗+w⃗)⋅det(ρ⃗1,...,β⃗,...,ρ⃗n)=vi⋅det(ρ⃗1,...,β⃗,...,ρ⃗n)+wi⋅det(ρ⃗1,...,β⃗,...,ρ⃗n)=det(ρ⃗1,...,vi⋅β⃗,...,ρ⃗n)+det(ρ⃗1,...,wi⋅β⃗,...,ρ⃗n)\begin{array}{rcl}\text{det}(\vec{\rho}_1,...,\vec{v}+\vec{w},...,\vec{\rho}_n)&=&\text{det}(\vec{\rho}_1,...,(v_i+w_i)\vec{\beta},...,\vec{\rho}_n)\\&=&(\vec{v}+\vec{w})\cdot\text{det}(\vec{\rho}_1,...,\vec{\beta},...,\vec{\rho}_n)\\&=&v_i\cdot\text{det}(\vec{\rho}_1,...,\vec{\beta},...,\vec{\rho}_n)+w_i\cdot\text{det}(\vec{\rho}_1,...,\vec{\beta},...,\vec{\rho}_n)\\&=&\text{det}(\vec{\rho}_1,...,v_i\cdot\vec{\beta},...,\vec{\rho}_n)+\text{det}(\vec{\rho}_1,...,w_i\cdot\vec{\beta},...,\vec{\rho}_n)\end{array}
Now add the vjρ⃗jv_j\vec{\rho}_j's to the first and wjρ⃗jw_j\vec{\rho}_j's to the second to recreate v⃗\vec{v} and w⃗\vec{w}, giving the desired expression.

Example 13.3

The determinant of
(1234)\begin{pmatrix}1&2\\3&4\end{pmatrix} can be split using multilinearity:
∣1234∣=∣0234∣+∣1034∣=∣1030∣+∣1004∣+∣0230∣+∣0204∣\begin{vmatrix}1&2\\3&4\end{vmatrix}=\begin{vmatrix}0&2\\3&4\end{vmatrix}+\begin{vmatrix}1&0\\3&4\end{vmatrix}=\begin{vmatrix}1&0\\3&0\end{vmatrix}+\begin{vmatrix}1&0\\0&4\end{vmatrix}+\begin{vmatrix}0&2\\3&0\end{vmatrix}+\begin{vmatrix}0&2\\0&4\end{vmatrix}
The left and right matrices are 00 becuase the second row is a scalar multiple of the first row.
=4⋅∣1001∣+2⋅3⋅∣0110∣=4\cdot\begin{vmatrix}1&0\\0&1\end{vmatrix}+2\cdot3\cdot\begin{vmatrix}0&1\\1&0\end{vmatrix}
We discuss evaluating these matrices below.

Example 13.4

The determinant can be reduced to a sum of determinants where each row has one element from the original.
∣123456789∣=∣100400700∣+∣100400080∣+⋯+∣003006080∣+∣003006009∣\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}=\begin{vmatrix}1&0&0\\4&0&0\\7&0&0\end{vmatrix}+\begin{vmatrix}1&0&0\\4&0&0\\0&8&0\end{vmatrix}+\cdots+\begin{vmatrix}0&0&3\\0&0&6\\0&8&0\end{vmatrix}+\begin{vmatrix}0&0&3\\0&0&6\\0&0&9\end{vmatrix}
If any two entries came from the same column, then one row is a multiple of another so the determinant is zero. This results in 66 determinants:
∣123456789∣=∣100050009∣+∣100006080∣+∣020400009∣+∣020006700∣+∣003400080∣+∣003050700∣=45⋅∣100010001∣+48⋅∣100001010∣+72⋅∣010100001∣+84⋅∣010001100∣+96⋅∣001100010∣+105⋅∣001010100∣\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}=\begin{vmatrix}1&0&0\\0&5&0\\0&0&9\end{vmatrix}+\begin{vmatrix}1&0&0\\0&0&6\\0&8&0\end{vmatrix}+\begin{vmatrix}0&2&0\\4&0&0\\0&0&9\end{vmatrix}+\begin{vmatrix}0&2&0\\0&0&6\\7&0&0\end{vmatrix}+\begin{vmatrix}0&0&3\\4&0&0\\0&8&0\end{vmatrix}+\begin{vmatrix}0&0&3\\0&5&0\\7&0&0\\\end{vmatrix}\\=45\cdot\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+48\cdot\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}+72\cdot\begin{vmatrix}0&1&0\\1&0&0\\0&0&1\end{vmatrix}+84\cdot\begin{vmatrix}0&1&0\\0&0&1\\1&0&0\end{vmatrix}+96\cdot\begin{vmatrix}0&0&1\\1&0&0\\0&1&0\end{vmatrix}+105\cdot\begin{vmatrix}0&0&1\\0&1&0\\1&0&0\\\end{vmatrix}
We discuss evaluating these matrices below.

Permutation Matrices

A permutation matrix is a matrix where every entry is 00 except for a single 11 in each row and column.
Define an nn-permutation as a function on the first nn integers ϕ:{1,...,n}→{1,...,n}\phi:\{1,...,n\}\to\{1,...,n\} that is bijective.
In other words, each of 1,...,n1,...,n in the output is associated with exactly one input.

Example 13.5

The 33-permutations are
ϕ1={1,2,3}ϕ2={1,3,2}ϕ3={2,1,3}ϕ4={2,3,1}ϕ5={3,1,2}ϕ6={3,2,1}\begin{array}{ccc}\phi_1=\{1,2,3\}&\phi_2=\{1,3,2\}&\phi_3=\{2,1,3\}\\\phi_4=\{2,3,1\}&\phi_5=\{3,1,2\}&\phi_6=\{3,2,1\}\end{array}

We denote the row matrix with all 00s except for a 11 in entry jj with ιj\iota_j (e.g. the four-wide ι2=(0200)\iota_2=\begin{pmatrix}0&2&0&0\end{pmatrix}). With this, we notate a permutation matrix such that a ϕ=⟨ϕ(1),...,ϕ(n)⟩\phi=\langle\phi(1),...,\phi(n)\rangle is associated with the matrix whose rows are ιϕ(1),...,ιϕ(n)\iota_{\phi(1)},...,\iota_{\phi(n)}
For example, for the 44-permutation ϕ:⟨2,4,3,1⟩\phi:\langle2,4,3,1\rangle, the matrix associated with it is
Pι=(ι2ι4ι3ι1)=(0100000100101000)P_\iota=\begin{pmatrix}\iota_2\\\iota_4\\\iota_3\\\iota_1\end{pmatrix}=\begin{pmatrix}0&1&0&0\\0&0&0&1\\0&0&1&0\\1&0&0&0\end{pmatrix}

Now we can define the permutation expansion for determinants:
∣t1,1t1,2⋯t1,nt2,1t2,2⋯t2,n⋮⋮⋱⋮tn,1tn,2⋯tn,n∣=t1,ϕ1(1)⋅t2,ϕ1(2)⋯tn,ϕ1(n)∣Pϕ1∣+t1,ϕ2(1)⋅t2,ϕ2(2)⋯tn,ϕ2(n)∣Pϕ2∣⋮+t1,ϕk(1)⋅t2,ϕk(2)⋯tn,ϕk(n)∣Pϕk∣\begin{vmatrix}t_{1,1}&t_{1,2}&\cdots&t_{1,n}\\t_{2,1}&t_{2,2}&\cdots&t_{2,n}\\\vdots&\vdots&\ddots&\vdots\\t_{n,1}&t_{n,2}&\cdots&t_{n,n}\end{vmatrix}=\begin{array}{c}t_{1,\phi_1(1)}\cdot t_{2,\phi_1(2)}\cdots t_{n,\phi_1(n)}|P_{\phi_1}|\\\\+t_{1,\phi_2(1)}\cdot t_{2,\phi_2(2)}\cdots t_{n,\phi_2(n)}|P_{\phi_2}|\\\vdots\\+t_{1,\phi_k(1)}\cdot t_{2,\phi_k(2)}\cdots t_{n,\phi_k(n)}|P_{\phi_k}|\end{array}
where ϕ1,...,ϕk\phi_1,...,\phi_k are the nn-permutations
In summation notation,
∣T∣=∑permutations ϕt1,ϕ1(1)⋅t2,ϕ(2)⋯tn,ϕ(n)∣Pϕ∣|T|=\sum_{\text{permutations }\phi}t_{1,\phi_1(1)}\cdot t_{2,\phi(2)}\cdots t_{n,\phi(n)}|P_{\phi}|
the sum of all permutations of ϕ\phi of the form t1,ϕ1(1)t2,ϕ(2)⋯tn,ϕ(n)∣Pϕ∣t_{1,\phi_1(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)}|P_{\phi}|

Example 13.6

Consider a 2×22\times2 matrix. There are two 22-permutations, ϕ1=⟨1,2⟩\phi_1=\langle1,2\rangle and ϕ2=⟨2,1⟩\phi_2=\langle2,1\rangle. The associated permutation matrices are
Pϕ1=(1001)Pϕ2=(0110)\begin{array}{cc}P_{\phi_1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}&P_{\phi_2}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\end{array}
So we get the expansion
∣abcd∣=ad⋅∣1001∣+bc⋅∣0110∣=ad⋅(1)+bc⋅(−1)=ad−bc\begin{array}{rcl}\begin{vmatrix}a&b\\c&d\end{vmatrix}&=&ad\cdot\begin{vmatrix}1&0\\0&1\end{vmatrix}+bc\cdot\begin{vmatrix}0&1\\1&0\end{vmatrix}\\&=&ad\cdot(1)+bc\cdot(-1)\\&=&ad-bc\end{array}
This gives the familiar formula for the determinant of the 2×22\times2 matrix.
Note that ∣0110∣=−1\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1 because it involves one row swap from II

Theorem: For each nn there is a n×nn\times n determinant function.
Theorem: The determinant of a matrix equals the determinant of its transpose.
This means statements about rows can be applied to the columns too; e.g. if row combinations don't change the determinant, column combinations don't either.
Also, a matrix is singular if two columns are equal, swapping columns changes the sign, and determinants are multilinear in their columns.

If TT is a lower triangular matrix, the determinant is still the product of the diaognal entries.
This is true because TTT^T has the same diagonals and is an upper triangular matrix, whose determinants are the product of the diagonal, and ∣T∣=∣TT∣|T|=|T^T|

Existance of Determinants

In a permutation ϕ=⟨...,k,...,j,...⟩\phi=\langle...,k,...,j,...\rangle or permutation matrix
Pϕ=(⋮ιk⋮ιj⋮)P_\phi=\begin{pmatrix}\vdots\\\iota_k\\\vdots\\\iota_j\\\vdots\end{pmatrix}
elements or rows such that k>jk>j or ιk>ιj\iota_k>\iota_j are in an inversion
ϕ=⟨3,2,1⟩\phi=\langle3,2,1\rangle has 3 inversions: 3>23>2, 2>12>1, 3>13>1

A row swap in a permutation matrix changes the parity of the number of inversions.

Proof

If the rows are adjacent, swapping the two won't affect the inversions of any other element, so it changes the number of inversions by 1.
If they are not, then swap them via a sequence, starting with bringing row kk up
(⋮ιϕ(j)ιϕ(j+1)⋮ιϕ(k)⋮)→ρk↔ρk−1 →ρk−1↔ρk−2 ⋯ →ρj+1↔ρj(⋮ιϕ(k)ιϕ(j)⋮ιϕ(k−1)⋮)\begin{pmatrix}\vdots\\\iota_{\phi(j)}\\\iota_{\phi(j+1)}\\\vdots\\\iota_{\phi(k)}\\\vdots\end{pmatrix}\xrightarrow{\rho_k\leftrightarrow\rho_{k-1}}\space\xrightarrow{\rho_{k-1}\leftrightarrow\rho_{k-2}}\space\cdots\space\xrightarrow{\rho_{j+1}\leftrightarrow\rho_j}\begin{pmatrix}\vdots\\\iota_{\phi(k)}\\\iota_{\phi(j)}\\\vdots\\\iota_{\phi(k-1)}\\\vdots\end{pmatrix}
then moving row jj down
→ρj+1↔ρj+2 →ρj+2↔ρj+3 ⋯ →ρk−1↔ρk(⋮ιϕ(k)ιϕ(j+1)⋮ιϕ(j)⋮)\xrightarrow{\rho_{j+1}\leftrightarrow\rho_{j+2}}\space\xrightarrow{\rho_{j+2}\leftrightarrow\rho_{j+3}}\space\cdots\space\xrightarrow{\rho_{k-1}\leftrightarrow\rho_k}\begin{pmatrix}\vdots\\\iota_{\phi(k)}\\\iota_{\phi(j+1)}\\\vdots\\\iota_{\phi(j)}\\\vdots\end{pmatrix}
The total number of swaps is (k−j)+(k−j−1)=2(k−j)−1(k-j)+(k-j-1)=2(k-j)-1, which is odd, so it changes the parity of the number of inversions

The signum of a permutation sgn(ϕ)\text{sgn}(\phi) is:
sgn(ϕ)={−1if number of inversions is odd+1if number of inversions is even\text{sgn}(\phi)=\begin{cases}-1&\text{if number of inversions is odd}\\+1&\text{if number of inversions is even}\end{cases}

If sgn(ϕ)=−1\text{sgn}(\phi)=-1, it takes an odd number of swaps to take it back to identity, and if sgn(ϕ)=1\text{sgn}(\phi)=1, it takes an even number (should be pretty intuitive)
⟹\implies∣Pϕ∣=sgn(ϕ)|P_\phi|=\text{sgn}(\phi) because a row swap changes the sign and ∣I∣=1|I|=1
Thus, the permutation expansion becomes
d(T)=∑permutations ϕt1,ϕ1(1)⋅t1,ϕ(2)⋯tn,ϕ(n)⋅sgn(ϕ)d(T)=\sum_{\text{permutations }\phi}t_{1,\phi_1(1)}\cdot t_{1,\phi(2)}\cdots t_{n,\phi(n)}\cdot\text{sgn}(\phi)
The signum function is clearly well-defined: just count the number of inversions.
So finally, we will show that this d(T)d(T) satisfies the conditions, proving that the determinant exists for all nn.

Proof

Condition (4) is easy: for II, the summation is all 00 except for the permutation which gives the product down the diagonal, which is 11.
For condition (3), suppose T→kρiT^T\xrightarrow{k\rho_i}\hat{T} and consider d(T^)d(\hat{T})
∑perm ϕt^1,ϕ(1)⋯t^i,ϕ(i)⋯t^n,ϕ(n)sgn(ϕ)=∑perm ϕt1,ϕ(1)⋯kti,ϕ(i)⋯tn,ϕ(n)sgn(ϕ)=k∑perm ϕt1,ϕ(1)⋯kti,ϕ(i)⋯tn,ϕ(n)sgn(ϕ)=k⋅d(T)\sum_{\text{perm }\phi}\hat{t}_{1,\phi(1)}\cdots\hat{t}_{i,\phi(i)}\cdots\hat{t}_{n,\phi(n)}\text{sgn}(\phi)=\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots kt_{i,\phi(i)}\cdots t_{n,\phi(n)}\text{sgn}(\phi)\\=k\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots kt_{i,\phi(i)}\cdots t_{n,\phi(n)}\text{sgn}(\phi)=k\cdot d(T)
leaves the desired equality
For condition (2), suppose T→ρi↔ρjT^T\xrightarrow{\rho_i\leftrightarrow\rho_j}\hat{T}. We must show d(T^)=−d(T)d(\hat{T})=-d(T)
t^i,ϕ(i)\hat{t}_{i,\phi(i)} and t^j,ϕ(j)\hat{t}_{j,\phi(j)} in each sum of the permutation expansion
∑perm ϕt^1,ϕ(1)⋯t^i,ϕ(i)⋯t^j,ϕ(j)⋯t^n,ϕ(n)sgn(ϕ)\sum_{\text{perm }\phi}\hat{t}_{1,\phi(1)}\cdots\hat{t}_{i,\phi(i)}\cdots\hat{t}_{j,\phi(j)}\cdots\hat{t}_{n,\phi(n)}\text{sgn}(\phi)
is taken by swapping the rows in TT, which as we have established before flips the sign. Thus,
∑perm ϕt^1,ϕ(1)⋯t^i,ϕ(i)⋯t^j,ϕ(j)⋯t^n,ϕ(n)⋅sgn(ϕ)=∑perm ϕt1,ϕ(1)⋯ti,ϕ(i)⋯tj,ϕ(j)⋯tn,ϕ(n)⋅(−sgn(ϕ))=−d(T)\sum_{\text{perm }\phi}\hat{t}_{1,\phi(1)}\cdots\hat{t}_{i,\phi(i)}\cdots\hat{t}_{j,\phi(j)}\cdots\hat{t}_{n,\phi(n)}\cdot\text{sgn}(\phi)=\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots t_{j,\phi(j)}\cdots t_{n,\phi(n)}\cdot(-\text{sgn}(\phi))\\ =-d(T)
For condition (1), suppose T→kρi+ρjT^T\xrightarrow{k\rho_i+\rho_j}\hat{T}
d(T^)=∑perm ϕt^1,ϕ(1)⋯ti,ϕ(i)⋯t^j,ϕ(j)⋯t^n,ϕ(n)sgn(ϕ)=∑perm ϕt1,ϕ(1)⋯ti,ϕ(i)⋯(kti,ϕ(j)+tj,ϕ(j))⋯tn,ϕ(n)sgn(ϕ)d(\hat{T})=\sum_{\text{perm }\phi}\hat{t}_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots\hat{t}_{j,\phi(j)}\cdots\hat{t}_{n,\phi(n)}\text{sgn}(\phi)\\ =\sum_{\text{perm }\phi} t_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots(kt_{i,\phi(j)}+t_{j,\phi(j)})\cdots t_{n,\phi(n)}\text{sgn}(\phi)
Distributing over addition and breaking into two summations:
=k⋅∑perm ϕt1,ϕ(1)⋯ti,ϕ(i)⋯ti,ϕ(j)⋯tn,ϕ(n)sgn(ϕ)+∑perm ϕt1,ϕ(1)⋯ti,ϕ(i)⋯tj,ϕ(j)⋯tn,ϕ(n)sgn(ϕ)=k\cdot\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots t_{i,\phi(j)}\cdots t_{n,\phi(n)}\text{sgn}(\phi)\\+\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots t_{j,\phi(j)}\cdots t_{n,\phi(n)}\text{sgn}(\phi)
See that the second term is d(T)d(T).
In the first term, the entry is ti,ϕ(j)t_{i,\phi(j)}, not tj,ϕ(j)t_{j,\phi(j)}. This sum represents the determinant of a matrix SS that is equal to TT except row jj of SS is row ii of TT, giving SS two copies of row ii. Thus, the first term is 00, making d(T^)=d(T)d(\hat{T})=d(T) as desired.

Thus, we have that for any nn, there exists a determinant function Mn×n→R\mathcal{M}_{n\times n}\to\mathbb{R}
Finally, we can show ∣T∣=∣TT∣|T|=|T^T| using the expansion.
∣T∣=∑perm ϕt1,ϕ(1)⋯ti,ϕ(i)⋯tj,ϕ(j)⋯tn,ϕ(n)sgn(ϕ)|T|=\sum_{\text{perm }\phi}t_{1,\phi(1)}\cdots t_{i,\phi(i)}\cdots t_{j,\phi(j)}\cdots t_{n,\phi(n)}\text{sgn}(\phi)
In TTT^T, the ta,ϕ(b)t_{a,\phi(b)}'s are all the same, since for ∣T∣|T| we have all ways to take one entry from each row and column of TT and TTT^T has all ways to take one entry from each column and row of TT. So, the only difference is in sgn(ϕ)\text{sgn}(\phi), but sgn(ϕ)=sgn(ϕ−1)\text{sgn}(\phi)=\text{sgn}(\phi^{-1}), so they are the same.

Determinants as Size Functions

A box or parallelepiped in Rn\mathbb{R}^n formed by ⟨v⃗1,...,v⃗n⟩\langle\vec{v}_1,...,\vec{v}_n\rangle is the set {t1v⃗1+⋯+tnv⃗n ∣ t1,...,tn∈[0,1]}\{t_1\vec{v}_1+\cdots+t_n\vec{v}_n\space|\space t_1,...,t_n\in[0,1]\}

A parallelepiped in R2\mathbb{R}^2

The determinant of the 2×22\times 2 matrix
∣x1x2y1y2∣\begin{vmatrix}x_1&x_2\\y_1&y_2\end{vmatrix}
represents the area of a parallelepiped in R2\mathbb{R}^2

Geometric Interpretations

Recall that the transpose does not change the determinant, so column operations are valid operations (just transposed row operations).
Also recall that scaling a column (or equivalently, a row) by kk scales the whole determinant by kk.
This makes sense, as it is analogous to scaling a side length of the box.

For the condition stating row combinations (or equivalent column combinations) does not change the determinant.

The base is the same, and the slant is different, but the height is the same, so the area remains the same.

Also, it is clear that the identity matrix has determinant of 11. A box made from (10)\begin{pmatrix}1\\0\end{pmatrix} and (01)\begin{pmatrix}0\\1\end{pmatrix} has area 11.

Swapping the vectors should negate the area. But area is positive, so the determinant sign of the determinant reflects the orientation or sense of the box.
This gives the right-hand rule in R3\mathbb{R}^3: do a "thumbs up" with your right hand and place it on the spanning plane so that your fingers curl from v⃗1\vec{v}_1 to v⃗2\vec{v}_2. Vectors on the side with the thumb define positive-sized boxes.

The determinant of the product of two matrices is the product of the determinants
∣TS∣=∣T∣∣S∣|TS|=|T||S|

Proof

First, suppose TT is singular and has no inverse. If TSTS is invertible, then there exists some MM such that (TS)M=T(SM)=I(TS)M=T(SM)=I, meaning TT must be invertible. The contrapositive says if TT is not invertible then neither is TSTS, so ∣T∣∣S∣=∣TS∣=0|T||S|=|TS|=0.
If TT is invertible, Then it is a product of elementary matrices T=E1E2⋯ErT=E_1E_2\cdots E_r. Showing ∣ES∣=∣E∣∣S∣|ES|=|E||S| for all matrices SS and elementary matrices EE proves the result.
For Mi(k)M_i(k), the matrix multiplying row ii by kk, we have ∣Mi(k)∣=k∣I∣=k|M_i(k)|=k|I|=k from condition three, but also ∣Mi(k)S∣=kS|M_i(k)S|=kS also from condition three. The case for the other two elementary matrices is similar.

From above, we can derive the determinant of the inverse:
1=∣I∣=∣TT−1∣=∣T∣∣T−1∣⟹∣T−1∣=1∣T∣1=|I|=|TT^{-1}|=|T||T^{-1}|\implies|T^{-1}|=\frac{1}{|T|}

The volume of a box is the absolute value of the determinant of a matrix with those vectors as columns.

Cramer's Rule

Recall that a linear system is equivalent to a linear vector equation.
x1+2x2=63x1+x2=8⟺x1(13)+x2(21)=(68)\begin{array}{ccc}\begin{array}{c}x_1+2x_2=6\\3x_1+x_2=8\end{array}&\iff&x_1\begin{pmatrix}1\\3\end{pmatrix}+x_2\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}6\\8\end{pmatrix}\end{array}
The geometric interpretation is to find what factors x1x_1 and x2x_2 must we scale the sides of the parallelogram so that it will fill the other vector.

Consider expanding only one side of the parallelogram, and compare the sizes of the shaded rectangles.

Together, we have
x1∣1231∣=∣x1⋅12x1⋅31∣=∣x1⋅1+x2⋅22x1⋅3+x2⋅11∣=∣6281∣x_1\begin{vmatrix}1&2\\3&1\end{vmatrix}=\begin{vmatrix}x_1\cdot1&2\\x_1\cdot3&1\end{vmatrix}=\begin{vmatrix}x_1\cdot1+x_2\cdot2&2\\x_1\cdot3+x_2\cdot1&1\end{vmatrix}=\begin{vmatrix}6&2\\8&1\end{vmatrix}
So dividing both sides,
x1=∣6281∣∣1231∣=−10−5=2x_1=\frac{\begin{vmatrix}6&2\\8&1\end{vmatrix}}{\begin{vmatrix}1&2\\3&1\end{vmatrix}}=\frac{-10}{-5}=2

This gives a new way to solve systems of equations.

Cramer's Rule
Let AA be an n×nn\times n matrix with nonzero determinant, let b⃗\vec{b} be an nn-tall column vector, consider the linear system Ax⃗=b⃗A\vec{x}=\vec{b}. For any i∈[1,...,n]i\in[1,...,n] let BiB_i be the matrix obtained by substituting b⃗\vec{b} for column ii in AA. Then the value of the ii-th unknown is xi=∣Bi∣/∣A∣x_i=|B_i|/|A|.
If the matrix has a zero determinant then the system has no solution.

Example 13.7

Solve the following system of equations
2x1+x2−x3=4x1+3x2=2x2−5x3=0\begin{array}{rcrcrcl}2x_1&+&x_2&-&x_3&=&4\\x_1&+&3x_2&&&=&2\\&&x_2&-&5x_3&=&0\end{array}

The corresponding matrix equation is
(21−113001−5)(x1x2x3)=(420)\begin{pmatrix}2&1&-1\\1&3&0\\0&1&-5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}4\\2\\0\end{pmatrix}
We can find
∣A∣=∣21−113001−5∣=−26∣B1∣=∣41−123001−5∣=−52\begin{array}{cc}|A|=\begin{vmatrix}2&1&-1\\1&3&0\\0&1&-5\end{vmatrix}=-26&|B_1|=\begin{vmatrix}4&1&-1\\2&3&0\\0&1&-5\end{vmatrix}\end{array}=-52
∣B2∣=∣24−112000−5∣=0∣B3∣=∣214132010∣=0\begin{array}{cc}|B_2|=\begin{vmatrix}2&4&-1\\1&2&0\\0&0&-5\end{vmatrix}=0&|B_3|=\begin{vmatrix}2&1&4\\1&3&2\\0&1&0\end{vmatrix}=0\end{array}
So the solutions are
(x1x2x3)=(∣B1∣ / ∣A∣∣B2∣ / ∣A∣∣B3∣ / ∣A∣)=(200)\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}|B_1|\space/\space|A|\\|B_2|\space/\space|A|\\|B_3|\space/\space|A|\end{pmatrix}=\begin{pmatrix}2\\0\\0\end{pmatrix}

Note that because this method requires taking the determinant, it is generally much slower to use Cramer's Rule for large matrices.

Laplace's Formula

Consider the permutation expansion
∣t1,1t1,2t1,3t2,1t2,2t2,3t3,1t3,2t3,3∣=t1,1t2,2t3,3∣100010001∣+t1,1t2,3t3,2∣100001010∣+t1,2t2,1t3,3∣010100001∣+t1,2t2,3t3,1∣010001100∣  +t1,3t2,1t3,2∣001100010∣+t1,3t2,2t3,1∣001010100∣\begin{array}{rl}\begin{vmatrix}t_{1,1}&t_{1,2}&t_{1,3}\\t_{2,1}&t_{2,2}&t_{2,3}\\t_{3,1}&t_{3,2}&t_{3,3}\end{vmatrix}=&t_{1,1}t_{2,2}t_{3,3}\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+t_{1,1}t_{2,3}t_{3,2}\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}\\&+t_{1,2}t_{2,1}t_{3,3}\begin{vmatrix}0&1&0\\1&0&0\\0&0&1\end{vmatrix}+t_{1,2}t_{2,3}t_{3,1}\begin{vmatrix}0&1&0\\0&0&1\\1&0&0\end{vmatrix}\\&\space\space+t_{1,3}t_{2,1}t_{3,2}\begin{vmatrix}0&0&1\\1&0&0\\0&1&0\end{vmatrix}+t_{1,3}t_{2,2}t_{3,1}\begin{vmatrix}0&0&1\\0&1&0\\1&0&0\end{vmatrix}\end{array}
Pick a row or column and factor out. Suppose we choose the first row
∣t1,1t1,2t1,3t2,1t2,2t2,3t3,1t3,2t3,3∣=t1,1[t2,2t3,3∣100010001∣+t2,3t3,2∣100001010∣]+t1,2[t2,1t3,3∣010100001∣+t2,3t3,1∣010001100∣]  +t1,3[t2,1t3,2∣001100010∣+t2,2t3,1∣001010100∣]\begin{array}{rl}\begin{vmatrix}t_{1,1}&t_{1,2}&t_{1,3}\\t_{2,1}&t_{2,2}&t_{2,3}\\t_{3,1}&t_{3,2}&t_{3,3}\end{vmatrix}=&t_{1,1}\left[t_{2,2}t_{3,3}\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+t_{2,3}t_{3,2}\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}\right]\\&+t_{1,2}\left[t_{2,1}t_{3,3}\begin{vmatrix}0&1&0\\1&0&0\\0&0&1\end{vmatrix}+t_{2,3}t_{3,1}\begin{vmatrix}0&1&0\\0&0&1\\1&0&0\end{vmatrix}\right]\\&\space\space+t_{1,3}\left[t_{2,1}t_{3,2}\begin{vmatrix}0&0&1\\1&0&0\\0&1&0\end{vmatrix}+t_{2,2}t_{3,1}\begin{vmatrix}0&0&1\\0&1&0\\1&0&0\end{vmatrix}\right]\end{array}
Using the property that a row swap changes the sign, swap the rows so that they match the first. This takes one swap for row 2 and two for row 3
∣t1,1t1,2t1,3t2,1t2,2t2,3t3,1t3,2t3,3∣=t1,1[t2,2t3,3∣100010001∣+t2,3t3,2∣100001010∣]−t1,2[t2,1t3,3∣100010001∣+t2,3t3,1∣100001010∣]  +t1,3[t2,1t3,2∣100010001∣+t2,2t3,1∣100001010∣]\begin{array}{rl}\begin{vmatrix}t_{1,1}&t_{1,2}&t_{1,3}\\t_{2,1}&t_{2,2}&t_{2,3}\\t_{3,1}&t_{3,2}&t_{3,3}\end{vmatrix}=&t_{1,1}\left[t_{2,2}t_{3,3}\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+t_{2,3}t_{3,2}\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}\right]\\&-t_{1,2}\left[t_{2,1}t_{3,3}\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+t_{2,3}t_{3,1}\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}\right]\\&\space\space+t_{1,3}\left[t_{2,1}t_{3,2}\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}+t_{2,2}t_{3,1}\begin{vmatrix}1&0&0\\0&0&1\\0&1&0\end{vmatrix}\right]\end{array}
The terms in the square brackets simplify to a 2×22\times2 determinants
=t1,1∣t2,2t2,3t3,2t3,3∣−t1,2∣t2,1t2,3t3,1t3,3∣+t1,3∣t2,1t2,2t3,1t3,2∣=t_{1,1}\begin{vmatrix}t_{2,2}&t_{2,3}\\t_{3,2}&t_{3,3}\end{vmatrix}-t_{1,2}\begin{vmatrix}t_{2,1}&t_{2,3}\\t_{3,1}&t_{3,3}\end{vmatrix}+t_{1,3}\begin{vmatrix}t_{2,1}&t_{2,2}\\t_{3,1}&t_{3,2}\end{vmatrix}

The i,ji,jminor for an n×nn\times n matrix TT is the (n−1)×(n−1)(n-1)\times(n-1) matrix formed by deleting row ii and column jj of TT. The i,ji,jcofactorTi,jT_{i,j} of TT is (−1)i+j(-1)^{i+j} times the determinant of the i,ji,j minor of TT

Example 13.8

For the matrix
S=(31254−170−3)S=\begin{pmatrix}3&1&2\\5&4&-1\\7&0&-3\end{pmatrix}
the 2,32,3 minor is
(3170)\begin{pmatrix}3&1\\7&0\end{pmatrix}
and the cofactor is
S2,3=(−1)2+3∣3170∣=7S_{2,3}=(-1)^{2+3}\begin{vmatrix}3&1\\7&0\end{vmatrix}=7

Laplace's formula finds the determinant of an n×nn\times n matrix by expanding by cofactors on any row ii or column jj
∣T∣=ti,1⋅Ti,1+ti,2⋅Ti,2+⋯+ti,n⋅Ti,n       =t1,j⋅T1,j+t2,j⋅T2,j+⋯+tn,j⋅Tn,j|T|=t_{i,1}\cdot T_{i,1}+t_{i,2}\cdot T_{i,2}+\cdots+t_{i,n}\cdot T_{i,n}\\\space\space\space\space\space\space\space=t_{1,j}\cdot T_{1,j}+t_{2,j}\cdot T_{2,j}+\cdots+t_{n,j}\cdot T_{n,j}

Example 13.9

Find the determinant
∣31254−170−3∣\begin{vmatrix}3&1&2\\5&4&-1\\7&0&-3\end{vmatrix}
by expanding along the second row
∣31254−170−3∣=−5∣120−3∣+4∣327−3∣−(−1)∣3170∣=−5(−3)+4(−23)+1(−7)=15−92−7=−84\begin{vmatrix}3&1&2\\5&4&-1\\7&0&-3\end{vmatrix}=-5\begin{vmatrix}1&2\\0&-3\end{vmatrix}+4\begin{vmatrix}3&2\\7&-3\end{vmatrix}-(-1)\begin{vmatrix}3&1\\7&0\end{vmatrix}\\=-5(-3)+4(-23)+1(-7)=15-92-7=-84

The matrix adjoint (or the classical adjoint or adjugate) of a square matrix TT is
adj(T)=(T1,1T2,1⋯Tn,1T1,2T2,2⋯Tn,2⋮T1,nT2,n⋯Tn,n)\text{adj}(T)=\begin{pmatrix}T_{1,1}&T_{2,1}&\cdots&T_{n,1}\\T_{1,2}&T_{2,2}&\cdots&T_{n,2}\\&\vdots\\T_{1,n}&T_{2,n}&\cdots&T_{n,n}\end{pmatrix}
note that the row ii column jj entry is Tj,iT_{j,i}, the j,ij,i cofactor.

Example 13.10

For the same matrix
S=(31254−170−3)S=\begin{pmatrix}3&1&2\\5&4&-1\\7&0&-3\end{pmatrix}
the matrix adjoint is
adj(S)=(−123−98−2313−2877)\text{adj}(S)=\begin{pmatrix}-12&3&-9\\8&-23&13\\-28&7&7\end{pmatrix}

For a square matrix TT, T⋅adj(T)=adj(T)⋅T=∣T∣⋅IT\cdot\text{adj}(T)=\text{adj}(T)\cdot T=|T|\cdot I. In other words,
(t1,1⋯t1,n⋮⋮tn,1⋯tn,n)(T1,1⋯T1,n⋮⋮Tn,1⋯Tn,n)=(∣T∣0⋯00∣T∣⋯0⋮00⋯∣T∣)\begin{pmatrix}t_{1,1}&\cdots&t_{1,n}\\\vdots&&\vdots\\t_{n,1}&\cdots&t_{n,n}\end{pmatrix}\begin{pmatrix}T_{1,1}&\cdots&T_{1,n}\\\vdots&&\vdots\\T_{n,1}&\cdots&T_{n,n}\end{pmatrix}=\begin{pmatrix}|T|&0&\cdots&0\\0&|T|&\cdots&0\\&\vdots\\0&0&\cdots&|T|\end{pmatrix}

Proof

Laplace's formula directly shows the diagonal entries are ∣T∣|T|.
For any off-diagonal entry, the multiplication gives
ti,1⋅Tk,1+ti,2⋅Tk,2+⋯+ti,n⋅Tk,n=0t_{i,1}\cdot T_{k,1}+t_{i,2}\cdot T_{k,2}+\cdots+t_{i,n}\cdot T_{k,n}=0
becuase it represents the expansion along row kk of a matrix with row ii equal to row kk, and a matrix with identical rows has determinant 00.

A result of this is that
T−1=1∣T∣adj(T)T^{-1}=\frac{1}{|T|}\text{adj}(T)