The determinant is a number associated with an 
matrix 
. It is defined as
where
-
is an element of in the first row of .^{1+j}M_{1j}^A)
is the cofactor associated with .
, the minor of the element 
, is the determinant of the 
matrix obtained by removing the 
row and 
-th column of .
In the case of 
, we say that the summation is a cofactor expansion along row 1. For example, the determinant of 
is
For any square matrix, the cofactor expansion along any row and any column results in the same determinant, i.e.
To prove this, we begin with the proof that the cofactor expansion along any row results in the same determinant, i.e. ^{i+j}M_{ij}^A)
for . Consider a matrix 
, which is obtained from by swapping row 
consecutively with rows above it 
times until it resides in row 1. According to property 8 (see below), we have
^{i-1}\vert B\vert=(-1)^{i-1}\sum_{j=1}^nb_{1j}(-1)^{1+j}M_{1j}^B\\=(-1)^{i-1}\sum_{j=1}^na_{ij}(-1)^{1+j}M_{ij}^A=\sum_{j=1}^na_{ij}(-1)^{i+j}M_{ij}^A)
According to property 10, 
and therefore, the cofactor expansion along any column also results in the same determinant. This concludes the proof.
In short, to calculate the determinant of an matrix, we can carry out the cofactor expansion along any row or column. If we expand along a row, we have 
. We then select any row to execute the summation. Conversely, if we expand along a column, we get 
.
The following are some useful properties of determinants:
-
, where is 
the identity matrix. If one of the diagonal elements of is 
, then 
.
- If the elements of one of the columns of are all zero,
.
- If is obtained from by multiplying the
-th row of by , 
.
- If is obtained from by swapping two rows or columns of , then
- If two rows or two columns of are the same, .
- The inverse of a matrix exists only if
.
- If
, then 
.
- If , then
. If , then 
.
- If is diagonal, then
.
Proof of property 1
We shall proof this property by induction.
For 
,
For 
,
(-1)^2M_{11}+(0)(-1)^3M_{12}=(1)(-1)^2(1)=1)
Let’s assume that for 
, 
. Then for 
,
I_{1n}+1=1)
We can repeat the above induction logic to prove that if one of the diagonal elements of is .
Proof of property 2
Again, we shall proof this property by induction.
For ,
For ,
(-1)^2M_{11}+c(0)(-1)^3M_{12}=c\vert ci_{22}\vert=c(c)=c^2)
Let’s assume that for , we have 
. Then for ,
I_{11}+c(0)I_{12}+\cdots+c(0)I_{1n}=cI_{11}=c\vert cI\vert_{n-1}=cc^{n-1}=c^n)
Proof of property 3
For 
, where 
, we have . For 
, the definition 
allows us to sum by row or by column. Suppose we sum by row, we have . Since we are allowed to choose any column to execute the summation, we can always select the column 
such that 
. Therefore, if the elements of one of the columns of are all zero.
Proof of property 4
Let’s suppose is obtained from by multiplying the -th row of by . If we expand 
and 
along row , cofactor 
is equal to cofactor 
. Therefore,
Proof of property 5
For a type I elementary matrix, 
transforms by swapping two rows of . So, 
due to property 8. Since 
is obtained from by swapping two rows of , we have 
according to property 1 and property 8, which implies that 
. Therefore, 
.
For a type II elementary matrix, 
due to property 4 and 
because of property 1. So, .
For a type III elementary matrix,
A_{pj}=\sum_{j=1}^na_{pj}A_{pj}+k\sum_{j=1}^na_{qj}A_{pj}=\vert A\vert+k\vert B\vert)
is computed by expanding along row . The equation 
means that when is computed by expanding along row , it has the same cofactor as when is computed by expanding along row . This implies that 
. Since the definition of the determinant of is 
, which in our case is equivalent to , we have 
. Thus 
, which according to property 9, gives:
Since 
,
according to eq5 and property 1.
Comparing eq5 and eq6, .
Proof of property 6
Case 1
If is singular, where , then 
is also singular according to property 12. So, .
Case 2
If is non-singular, it can always be expressed as a product of elementary matrices: 
. So,
\vert)
Since property 5 states that 
,
Similarly, \vert=\vert\varepsilon_1\vert\vert\varepsilon_2\vert\cdots\vert\varepsilon_k\vert)
. Substitute this in the above equation, .
Proof of property 7
Using property 6 and then property 2,
Proof of property 8
We shall proof this property by induction. 
is the trivial case, where 
is the rank of a square matrix.
For , let 
and 
, which is obtained from by swapping two adjacent rows. Furthermore, let ^{1+j}m_{1j}^A)
and ^{1+j}m_{1j}^B)
. Clearly,
=-\vert A\vert)
Let’s assume that for 
, when two adjacent rows are swapped. For 
, we have:
Case 1: Suppose that the first row of is not swapped when making .
^{1+j}M_{1j}^B=\sum_{j=1}^na_{1j}(-1)^{1+j}M_{1j}^B)
is the determinant of a rank matrix, which is the same as except for two adjacent rows being swapped. Therefore, 
and ^{1+j}M_{1j}^A=-\vert A\vert)
.
Case 2: If the first two rows of are swapped when making ,
We have ^{1+j}M_{1j}^A)
and ^{1+k}M_{1k}^B=\sum_{k=1}^na_{2k}(-1)^{1+k}M_{1k}^B)
. The minors and 
can be expressed as
\sum_{k=1}^{j-1}a_{2k}(-1)^{1+k}\vert A_{kj}\vert+\sum_{k=j+1}^na_{2k}(-1)^{k}\vert A_{jk}\vert)
\sum_{j=1}^{k-1}a_{1j}(-1)^{1+j}\vert A_{jk}\vert+\sum_{j=k+1}^na_{1j}(-1)^{j}\vert A_{kj}\vert)
where 
is with the first two rows, and the -th and -th columns removed.
Therefore,
^{1+j}\biggr\[(1-\delta_{1j})\sum_{k=1}^{j-1}a_{2k}(-1)^{1+k}\vert A_{kj}\vert+\sum_{k=j+1}^na_{2k}(-1)^{k}\vert A_{jk}\vert\biggr\])
^{1+k}\biggr\[(1-\delta_{1k})\sum_{j=1}^{k-1}a_{1j}(-1)^{1+j}\vert A_{jk}\vert+\sum_{j=k+1}^na_{1j}(-1)^{j}\vert A_{kj}\vert\biggr\])
For any pair of values of and , where 
, the terms in are ^{1+j}\sum_{k=1}^{j-1}a_{2k}(-1)^{1+k}\vert A_{kj}\vert)
, which differ from the terms in , i.e. ^{1+k}\sum_{j=k+1}^{n}a_{1j}(-1)^{j}\vert A_{kj}\vert)
, by a factor of -1. Similarly, for any pair of values of and , where 
, the terms in are ^{1+j}\sum_{k=j+1}^{n}a_{2k}(-1)^{k}\vert A_{jk}\vert)
, which again differ from the terms in , i.e. ^{1+k}\sum_{j=1}^{k-1}a_{1j}(-1)^{1+j}\vert A_{jk}\vert)
, by a factor of -1. Since all terms in differ from all corresponding terms in by a factor of -1, .
In general, the swapping of any two rows and 
of , where 
, is equivalent to the swapping of -1)
adjacent rows of , with each swap changing by a factor of -1. Therefore,
^{2(q-p)-1}\vert A\vert=[(-1)^2]^{p-q}(-1)^{-1}\vert A\vert=-\vert A\vert)
Finally, the swapping of any two columns is proven in a similar way.
Proof of property 9
Consider the swapping of two equal rows of to form , resulting in 
and 
. However, property 8 states that if any two rows of are swapped. Therefore, 
if two rows of are equal. The same logic applies to proving if there are two equal columns of .
Proof of property 10
Case 1:
If , then 
according to property 13. So, 
.
Case 2: 
Let’s first consider elementary matrices 
. A type I elementary matrix is symmetrical about its diagonal, while a type II elementary matrix has one diagonal element equal to . Therefore, 
and thus 
for type I or II elementary matrices. A type III elementary matrix is an identity matrix with one of the non-diagonal elements replaced by a constant . Therefore, if is a type III elementary matrix, then 
is also one. According to eq6, 
for a type III elementary matrix. Hence, 
for all elementary matrices.
Next, consider an invertible matrix , which (as proven in the previous article) can be expressed as . Thus, ^T=\varepsilon_k^T\cdots\varepsilon_2^T\varepsilon_1^T)
(see Q&A in the proof of property 13). According to property 5,
and
Therefore, .
Proof of property 11
We have 
, or in terms of matrix components:
_{qk}a_{kp}=\delta_{pq}=\frac{\vert A\vert}{\vert A\vert}\delta_{pq}\;\;\;\;\;\;\;\;7)
Consider the matrix that is obtained from the matrix by replacing the -th column of with the -th column, i.e. 
for 
and 
. According to property 9, 
because has two equal columns. Furthermore, cofactor 
is equal to cofactor 
for . Therefore,
When 
, the last summation in eq8 becomes
Combining eq8 and eq9, we have 
, which when substituted in eq7 gives:
_{qk}a_{kp}=\sum_{k=1}^n\frac{A_{kq}}{\vert A\vert}a_{kp})
Therefore, _{qk}=\frac{A_{kq}}{\vert A\vert})
, which implies that the inverse of a matrix is undefined if . In other words, the inverse of a matrix is undefined if . We call such a matrix, a singular matrix, and a matrix with an associated inverse, a non-singular matrix.
Proof of property 12
We shall prove by contradiction. According to property 11, has no inverse if . If has no inverse and has an inverse, then ^{-1}\]=I)
. This implies that has an inverse 
, where ^{-1})
, which contradicts the initial assumption that has no inverse. Therefore, if has no inverse, then must also have no inverse.
Proof of property 13
Question
Show that ^T=\cdots C^TB^TA^T)
.
Answer
^T=B^TA^T)
because
^T_{ij}=(AB)_{ji}=\sum_{k=1}^na_{jk}b_{ki}=\sum_{k=1}^n(A^T)_{kj}(B^T)_{ik}\\=\sum_{k=1}^n(B^T)_{ik}(A^T)_{kj}=(B^TA^T)_{ij})
^T=C^TB^TA^T)
because
^T_{ij}=(ABC)_{ji}=\sum_{l=1}^n\sum_{k=1}^na_{jk}b_{kl}c_{li}=\sum_{l=1}^n\sum_{k=1}^n(A^T)_{kj}(B^T)_{lk}(C^T)_{il}\\=\sum_{l=1}^n\sum_{k=1}^n(C^T)_{il}(B^T)_{lk}(A^T)_{kj}=(C^TB^TA^T)_{ij})
which can be extended to .
Using the identity in the above Q&A, ^TA^T=(AA^{-1})^T)
. If is invertible, then ^TA^T=(AA^{-1})^T=I=(A^{-1}A)^T=A^T(A^{-1})^T)
. This implies that ^T)
is the inverse of 
and therefore that is invertible if is invertible.
The last part shall be proven by contradiction. Suppose is singular and is non-singular, there would be a matrix such that 
, Furthermore, ^T=AB^T)
, which implies that 
. This contradicts our initial assumption that is singular. Therefore, if is singular, must also be singular.
Proof of property 14
We shall proof this property by induction. For ,
Let’s assume that 
for . Then for 
, the cofactor expansion along the first row is
