preface

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Self-introduction ଘ(੭, ᵕ)੭

Nickname: Haihong

Tag: programmer monkey | C++ contestant | student

Introduction: because of C language to get acquainted with programming, then transferred to the computer major, had the honor to get some national awards, provincial awards… Has been confirmed. Currently learning C++/Linux/Python

Learning experience: solid foundation + more notes + more code + more thinking + learn English well!

 

Machine learning little White stage

The article is only used as my own study notes for the establishment of knowledge system and review

Know what is, know why!

Warm prompt

Because of compiler compatibility

Some latex grammatical mathematical formulas are displayed incorrectly or incorrectly

Linear Algebra for Machine Learning (3) : Properties of determinants

1.5 Properties of the determinant

Transpose determinant

The determinants of order N D: D= epsilon a11a12… a1na21a22… a2n…… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Aij =ajia_{ij}=a_{ji}aij=aji DT=∣a11a21… an1a12a22… an2…… a1na2n… Ann ∣D^T=\begin{vmatrix} a_{11} &a_ {21} &… & a_{n1}\\ a_{12} & a_{22} & … &a_{n2}\\ . & . & & . \\ . & . & & . \\ a_{1n} & a_{2n} &… & a_ {nn} {vmatrix} \ \ \ end DT = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a12. J. a1na21a22.. a2n……… an1an2.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

The determinant DTD^TDT is called the transpose determinant of the determinant DDD

The nature of the 1

content

The determinant is the same as its transpose determinant

prove

Set D = ∣ a11a12… a1na21a22… a2n…… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann given given given given given given given given given DTDT is the transpose determinant of DDD

Again a DT = ∣ b11b12… b1nb21b22… b2n…… bn1bn2… BNN ∣D^T=\begin{vmatrix} b_{11} &b_ {12} &… & b_{1n}\\ b_{21} & b_{22} & … &b_{2n}\\ . & . & & . \\ . & . & & . \\ b_{n1} & b_{n2} &… & b_ {nn} {vmatrix} \ \ \ end DT = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ b11b21. J. bn1b12b22.. bn2……… b1nb2n.. BNN ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

And because we know DT is equal to ∣a11a21 an1a12a22… an2…… a1na2n… Ann ∣D^T=\begin{vmatrix} a_{11} &a_ {21} &… & a_{n1}\\ a_{12} & a_{22} & … &a_{n2}\\ . & . & & . \\ . & . & & . \\ a_{1n} & a_{2n} &… & a_ {nn} {vmatrix} \ \ \ end DT = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a12. J. a1na21a22.. a2n……… an1an2.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

So bij=ajib_{ij}=a_{ji}bij=aji

Introduce DT = ∣ b11b12… b1nb21b22… b2n…… bn1bn2… BNN ∣D^T=\begin{vmatrix} b_{11} &b_ {12} &… & b_{1n}\\ b_{21} & b_{22} & … &b_{2n}\\ . & . & & . \\ . & . & & . \\ b_{n1} & b_{n2} &… & b_ {nn} {vmatrix} \ \ \ end DT = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ b11b21. J. bn1b12b22.. bn2……… b1nb2n.. BNN ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∑ (1 -) tb1p1b2p2… bnpn\sum(-1)^tb_{1p_1}b_{2p_2}… B_ {np_n} ∑ (1 -) tb1p1b2p2… BNPN = ∑ (1 -) tap11ap22… apnn\sum(-1)^ta_{p_11}a_{p_22}… A_ {p_nn} ∑ (1 -) tap11ap22… Apnn (using bij = ajib_ {ij} = a_ {ji} bij = aji)

And because


∑ ( – 1 ) t a 1 p 1 a 2 p 2 . . . a n p n = ∑ ( – 1 ) t a p 1 1 a p 2 2 . . . a p n n \sum(-1)^ta_{1p_1}a_{2p_2}… a_{np_n}=\sum(-1)^ta_{p_11}a_{p_22}… a_{p_nn}

so

Proof done!

The nature of the 2

content

Swap two rows (columns) of the determinant, the determinant changes sign

prove

Let’s say the determinant of order N D D= epsilon a11a12… a1na21a22… a2n…… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Swap row I and row j

Have to D1D_1D1

We set up


D 1 = ∣ b 11 b 12 . . . b 1 n b 21 b 22 . . . b 2 n . . . . . . b n 1 b n 2 . . . b n n ∣ D_1=\begin{vmatrix} b_{11} & b_{12} &… & b_{1n}\\ b_{21} & b_{22} & … &b_{2n}\\ . & . & & . \\ . & . & & . \\ b_{n1} & b_{n2} &… & b_{nn}\\ \end{vmatrix}

When k indicates I, jk \ neq I, jk  = I, j, a BKP = akpb_ KP} {= a_ KP} {BKP = akp when k = I, jk = I, jk = I, j, IP has BJP = ajpb_ {} = a_ jp} {BJP = ajp, BJP = aipb_ jp} {= a_} {IP BJP = aip

In simple terms

  • If k≠ I,jk\neq I,jk= I,j, that is, the two rows that do not belong to the exchange, b and A are the complete correspondence;
  • When k is equal to I,jk is equal to I,jk is equal to I,j, those are the two rows that swap, and the rows between B and A are the opposite, the columns are the same


D 1 = ∣ b 11 b 12 . . . b 1 n b 21 b 22 . . . b 2 n . . . . . . b n 1 b n 2 . . . b n n ∣ = ∑ ( – 1 ) t b 1 p 1 . . . b i p i . . . b j p j . . . b n p n = ∑ ( – 1 ) t a 1 p 1 . . . a j p i . . . a i p j . . . a n p n D_1=\begin{vmatrix} b_{11} & b_{12} &… & b_{1n}\\ b_{21} & b_{22} & … &b_{2n}\\ . & . & & . \\ . & . & & . \\ b_{n1} & b_{n2} &… & b_{nn}\\ \end{vmatrix}=\sum(-1)^tb_{1p_1}… b_{ip_i}… b_{jp_j}… b_{np_n}=\sum(-1)^ta_{1p_1}… a_{jp_i}… a_{ip_j}… a_{np_n}

with


D = ∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . a j p j . . . a n p n D=\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}

Comparison shows that only row coordinates are swapped

From 1… i… j… N becomes 1… j… i… n

And obviously, just like swapping any two elements in a complete permutation, the parity changes which is the inverse number plus 1 or minus 1

It’s just that in the determinant it’s equivalent to a change in parity, which is multiplied by a minus 1

so


D 1 = – D D_1=-D

Proof done!

The nature of the three

content

Multiplying all elements of a row (column) by the same number k is equal to multiplying the determinant by the number k

prove

Let’s say the determinant D= epsilon a11a12… a1na21a22… a2n…… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21. J. an1a12a22.. an2……… a1na2n.. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

If I multiply all of the entries in row I times k, I get 1, 2, 3


D 1 = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . a i 1 ∗ k a i 2 ∗ k . a i n ∗ k . . . a n 1 a n 2 . . . a n n ∣ D_1=\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ a_{i1}*k & a_{i2}*k& . & a_{in}*k\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}

reduction


D 1 = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . a i 1 ∗ k a i 2 ∗ k . a i n ∗ k . . . a n 1 a n 2 . . . a n n ∣ = ∑ ( – 1 ) t a 1 p 1 a 2 p 2 . . . k ∗ a i p i . . . a n p n = k ∗ ∑ ( – 1 ) t a 1 p 1 a 2 p 2 . . . a i p i . . . a n p n D_1=\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ a_{21} & a_{22} & … &a_{2n}\\ . & . & & . \\ a_{i1}*k & a_{i2}*k& . & a_{in}*k\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}=\sum(-1)^ta_{1p_1}a_{2p_2}… k*a_{ip_i}… a_{np_n}=k*\sum(-1)^ta_{1p_1}a_{2p_2}… a_{ip_i}… a_{np_n}

Proof done!

The same is true for columns

The nature of the 4

content

The determinant is equal to 0 if there are two rows (columns) of elements in a proportional column

prove

Let’s say the determinant D= epsilon a11a12… a1n… ai1ai2… ain… aj1aj2… ajn… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ . & . & & . \\ a_{i1} & a_{i2} & … &a_{in}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11. Ai1, aj1. An1a12. Ai2. Aj2. An2… A1n. Ain. Ajn. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Where the element in row I is proportional to the element in row J, that is, AJP =k∗ AjPA_ {jp}= K *a_{jp}ajp=k∗ajp


D = ∣ a 11 a 12 . . . a 1 n . . . a i 1 a i 2 . . . a i n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ = ∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . a j p j . . . a n p n D=\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ a_{i1} & a_{i2} & … &a_{in}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix} =\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}

Because ajp = k ∗ ajpa_ jp} {= k * a_ jp} {ajp = k ∗ ajp

so


∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . a j p j . . . a n p n = ∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . k ∗ a i p i . . . a n p n = k ∗ ∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . a i p i . . . a n p n = D \sum(-1)^ta_{1p_1}… a_{ip_i}… a_{jp_j}… a_{np_n}=\sum(-1)^ta_{1p_1}… a_{ip_i}… k*a_{ip_i}… a_{np_n}=k*\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{ip_i}… a_{np_n}=D

If I swap row I and row j

The determinant DDD still doesn’t change

Because D is going to be the same

K ∗ ∑ (1 -) ta1p1… aipi… aipi… anpn=Dk*\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{ip_i}… A_ {np_n} = Dk ∗ ∑ (1 -) ta1p1… aipi… aipi… Anpn is equal to D.

But because any two rows, when you swap them, you have to change the sign to -d

In conclusion, there are


D = – D D=-D

get


D = 0 D=0

Proof done!

Nature of the five

content

If the elements of a row (column) are the sum of two numbers, for example, the ith column is the sum of two numbers

D = ∣ a11a12… (a1i+b1i)… a1na21a22… (a2i+b2i)… a2n……… an1an2… (ani+bni)… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… &(a_{1i}+b_{1i})&… & a_{1n}\\ a_{21} & a_{22} & … &(a_{2i}+b_{2i})&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &(a_{ni}+b_{ni})&… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11a21… an1a12a22… an2……… (a1i+b1i)(a2i+b2i)… (ani+bni)……… A1na2nann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ criterion


D = ∣ a 11 a 12 . . . a 1 i . . . a 1 n a 21 a 22 . . . a 2 i . . . a 2 n . . . . . . . . . a n 1 a n 2 . . . a n i . . . a n n ∣ + ∣ a 11 a 12 . . . b 1 i . . . a 1 n a 21 a 22 . . . b 2 i . . . a 2 n . . . . . . . . . a n 1 a n 2 . . . b n i . . . a n n ∣ D=\begin{vmatrix} a_{11} & a_{12} &… &a_{1i}&… & a_{1n}\\ a_{21} & a_{22} & … &a_{2i}&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &a_{ni}&… & a_{nn}\\ \end{vmatrix} +\begin{vmatrix} a_{11} & a_{12} &… &b_{1i}&… & a_{1n}\\ a_{21} & a_{22} & … &b_{2i}&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &b_{ni}&… & a_{nn}\\ \end{vmatrix}

prove


D = ∣ a 11 a 12 . . . ( a 1 i + b 1 i ) . . . a 1 n a 21 a 22 . . . ( a 2 i + b 2 i ) . . . a 2 n . . . . . . . . . a n 1 a n 2 . . . ( a n i + b n i ) . . . a n n ∣ = ∑ ( – 1 ) t a 1 p 1 . . . ( a i p i + b i p i ) . . . a n p n = ∑ ( – 1 ) t ( a 1 p 1 . . . a i p i . . . a n p n + a 1 p 1 . . . b i p i . . . a n p n ) = ∑ ( – 1 ) t a 1 p 1 . . . a i p i . . . a n p n + ∑ ( – 1 ) t a 1 p 1 . . . b i p i . . . a n p n = ∣ a 11 a 12 . . . a 1 i . . . a 1 n a 21 a 22 . . . a 2 i . . . a 2 n . . . . . . . . . a n 1 a n 2 . . . a n i . . . a n n ∣ + ∣ a 11 a 12 . . . b 1 i . . . a 1 n a 21 a 22 . . . b 2 i . . . a 2 n . . . . . . . . . a n 1 a n 2 . . . b n i . . . a n n ∣ D=\begin{vmatrix} a_{11} & a_{12} &… &(a_{1i}+b_{1i})&… & a_{1n}\\ a_{21} & a_{22} & … &(a_{2i}+b_{2i})&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &(a_{ni}+b_{ni})&… & a_{nn}\\ \end{vmatrix} =\sum(-1)^ta_{1p_1}… (a_{ip_i}+b_{ip_i})… a_{np_n} \\=\sum(-1)^t(a_{1p_1}… a_{ip_i}… a_{np_n}+a_{1p_1}… b_{ip_i}… a_{np_n})\\=\sum(-1)^ta_{1p_1}… a_{ip_i}… a_{np_n}+\sum(-1)^ta_{1p_1}… b_{ip_i}… a_{np_n}\\=\begin{vmatrix} a_{11} & a_{12} &… &a_{1i}&… & a_{1n}\\ a_{21} & a_{22} & … &a_{2i}&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &a_{ni}&… & a_{nn}\\ \end{vmatrix} +\begin{vmatrix} a_{11} & a_{12} &… &b_{1i}&… & a_{1n}\\ a_{21} & a_{22} & … &b_{2i}&… &a_{2n}\\ . & . & & . \\ . & . & & . \\ . & . & & . \\ a_{n1} & a_{n2} &… &b_{ni}&… & a_{nn}\\ \end{vmatrix}

Proof done!

The nature of the 6

content

You multiply each element of one row of the determinant by the same number and then add it to the corresponding element of the other row, the same determinant

prove

Let’s say the determinant D= epsilon a11a12… a1n… ai1ai2… ain… aj1aj2… ajn… an1an2… Ann ∣D=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ . & . & & . \\ a_{i1} & a_{i2} & … &a_{in}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} D = \ \ \ end ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11. Ai1, aj1. An1a12. Ai2. Aj2. An2… A1n. Ain. Ajn. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

You multiply the JTH row times k, add it to the ith row, and you get


D 1 = ∣ a 11 a 12 . . . a 1 n . . . a i 1 + k ∗ a j 1 a i 2 + k ∗ a j 2 . . . a i n + k ∗ a j n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ D_1=\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ a_{i1}+k*a_{j1} & a_{i2}+k*a_{j2} & … &a_{in}+k*a_{jn}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}

That’s property 5


D 1 = ∣ a 11 a 12 . . . a 1 n . . . a i 1 + k ∗ a j 1 a i 2 + k ∗ a j 2 . . . a i n + k ∗ a j n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ = ∣ a 11 a 12 . . . a 1 n . . . a i 1 a i 2 . . . a i n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ + ∣ a 11 a 12 . . . a 1 n . . . k ∗ a j 1 k ∗ a j 2 . . . k ∗ a j n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ D_1=\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ a_{i1}+k*a_{j1} & a_{i2}+k*a_{j2} & … &a_{in}+k*a_{jn}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix} =\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ a_{i1} & a_{i2} & … &a_{in}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix} +\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ k*a_{j1} & k*a_{j2} & … &k*a_{jn}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix}

That’s property 4


∣ a 11 a 12 . . . a 1 n . . . k ∗ a j 1 k ∗ a j 2 . . . k ∗ a j n . . . a j 1 a j 2 . . . a j n . . . a n 1 a n 2 . . . a n n ∣ = 0 \begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ k*a_{j1} & k*a_{j2} & … &k*a_{jn}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix} =0

So the D1 = ∣ a11a12… a1n… Ai1 + k ∗ ∗ aj1ai2 + k aj2… Ain + k ∗ ajn… aj1aj2… ajn… an1an2… Ann ∣ = ∣ a11a12… a1n… ai1ai2… ain… aj1aj2… ajn… an1an2… Ann ∣=DD_1=\begin{vmatrix} a_{11} &a_ {12} &… & a_{1n}\\ . & . & & . \\ a_{i1}+k*a_{j1} & a_{i2}+k*a_{j2} & … &a_{in}+k*a_{jn}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_{nn}\\ \end{vmatrix} =\begin{vmatrix} a_{11} & a_{12} &… & a_{1n}\\ . & . & & . \\ a_{i1} & a_{i2} & … &a_{in}\\ . & . & & . \\ a_{j1} & a_{j2} & … &a_{jn}\\ . & . & & . \\ a_{n1} & a_{n2} &… & a_ {nn} {vmatrix} \ \ \ end = DD1 = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11. Ai1 + k ∗ aj1, aj1. An1a12. Ai2 + k ∗ aj2. Aj2. An2… A1n. Ain + k ∗ ajn. Ajn. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a11. Ai1, aj1. An1a12. Ai2. Aj2. An2… A1n. Ain. Ajn. Ann ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = D

Proof done!

conclusion

Description:

  • Refer to textbook “linear algebra” fifth edition tongji University mathematics department
  • With the book concept explanation combined with some of their own understanding and thinking

The essay is just a study note, recording a process from 0 to 1

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