A matrix is a set of numbers
arranged in m rows and n
columns.
Some definition associated with matrices
Special matrices:
Hermitian Matrix
Skew-Hermitian matrix
Gauss-Jordan Elimination Method:
A method of solving a linear system of equations. This is done by transforming the
system's augmented matrix into reduced row-echelon form by means of elementary
operations.
The following row operations on the augmented matrix of a system produce the
augmented matrix of an equivalent system, i.e., a system with the same solution as the
original one. The Gauss-Jordan elimination method to solve a system of linear
equations is described in the following steps.
1- Interchange any two rows. 𝑅𝑖 ↔ 𝑅𝑗 means: Interchange row 𝑖 and row 𝑗.
2- Multiply each element of a row by a nonzero constant. 𝛼𝑅𝑖 means: Replace row 𝑖
with 𝛼 (𝛼 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) times row 𝑖
3- Replace a row by the sum of itself and a constant multiple of another row of the
matrix. 𝑅𝑖 + 𝛼𝑅𝑗 means: Replace row 𝑖 with the sum of row i and α times row 𝑗.
Rank: Rank of Matrix is: The maximum number of linearly independent rows in
a matrix A is called the row rank of A.
e.g:
If the system equations is Ax=B then the solution can be have one of the following cases
Where n is the numbers of variables
e.g.:
Example 1: Solve the following system by using the Gauss-Jordan elimination method.
𝟐𝒙 + 𝟑𝒚 + 𝟓𝒛 = 𝟖 ,
𝒙+𝒚+𝒛=𝟓 ,
𝟒𝒙 + 𝟓𝒛 = 𝟐
2 3
Solution: The augmented matrix of the system is the following. 1 1
4 0
2 3
. 1 1
4 0
5 8
1 5
5 2
𝑅2 ⟷𝑅1
1
2
4
𝑅3 =𝑅3 −4𝑅1
1
𝑅3 = 𝑅3
13
1 1
0 1
0 0
1
3
0
1 5
5 8
5 2
1
0
0
1
1
−4
1 5
3 −2
1 −2
𝑅2 =𝑅2 −2𝑅1
1 5
3 −2
1 −18
𝑅1 =𝑅1 −𝑅3
𝑅2 =𝑅2 −3𝑅3
1
0
4
1 1 5
1 3 −2
0 5 2
𝑅3 =𝑅3 +4𝑅2
1 1
0 1
0 0
5 8
1 5
5 2
0 7
0 4
1 −2
1 1 1 5
0 1 3 −2
0 0 13 −26
𝑅1 =𝑅1 −𝑅2
1 0 0 3
0 1 0 4
0 0 1 −2
⟹ 𝑹𝒂𝒏𝒌 𝑨 = 𝑹𝒂𝒏𝒌 𝑨 𝑩 = 𝒏 From this final matrix, we can read the solution of the
system. It is 𝒙 = 𝟑, 𝒚 = 𝟒, 𝒛 = −𝟐.
Example 2: Solve the following system by using the Gauss-Jordan elimination method.
𝒙 + 𝒚 + 𝟐𝒛 = 𝟏 ,
𝟐𝒙 − 𝒚 + 𝟐𝒛 = 𝟑 ,
𝟒𝒙 + 𝒚 + 𝟔𝒛 = 𝟓
1 1
Solution: The augmented matrix of the system is the following. 2 −1
4 1
1
2
4
1 2 1
−1 2 3
1 6 5
𝑅2 =𝑅2 −2𝑅1
𝑅3 =𝑅3 −4𝑅1
𝑹𝒂𝒏𝒌 𝑨 = 𝑹𝒂𝒏𝒌 𝑨 𝑩 < 𝒏
−3𝑦 − 2𝑧 = 1 ⇒ 𝑦 =
1+2𝑧
−3
1
0
0
1
2 1
−3 −2 1
−3 −2 1
𝑅3 =𝑅3 −𝑅2
We obtain infinite solutions
, 𝑥 + 𝑦 + 2𝑧 = 1 ⇒ 𝑥 =
4−4𝑧
3
1
0
0
1
−3
0
2 1
1 3
5 4
2 1
−2 1
0 0
Example 3: Solve the following system by using the elementary row operation.
𝟑𝒙 + 𝟐𝒚 − 𝒛 = 𝟕 ,
𝒙−𝒚+𝒛=𝟒 ,
−𝒙 − 𝟒𝒚 + 𝟑𝒛 = 𝟐
1
Solution: The augmented matrix of the system is the following. 3
−1
1
3
−1
−1 1 4
2 −1 7
−4 3 2
𝑅2 =𝑅2 −3𝑅1
𝑅3 =𝑅3 +𝑅1
1 −1 1 4
0 5 −4 −5
0 −5 4 6
𝑅3 =𝑅3 +𝑅1
−1
2
−4
1
0
0
1 4
−1 7
3 2
−1 1 4
5 −4 −5
0
0 1
The matrix has now echelon form, and there is a pivot in the last column. Therefore, the linear
system has no solutions. In fact, the
𝑹𝒂𝒏𝒌 𝑨 ≠ 𝑹𝒂𝒏𝒌 𝑨 𝑩
which clearly has no solutions.
Eigenvalues and Eigenvectors
A (non-zero) vector V of dimension N is an eigenvector of a square (n×m) matrix A if it
satisfies the linear equation 𝐴 𝑉 = 𝜆 𝑉 , where {\displaystyle \lambda } 𝜆 is a scalar,
termed the eigenvalue corresponding to V.
That is, the eigenvectors are the vectors that the linear transformation A merely
elongates or shrinks, and the amount that they elongate/shrink by is the eigenvalue.
The above equation is called the eigenvalue equation or the eigenvalue problem.
This yields an equation for the eigenvalues P(λ)=det (A-λI)=0
We call P(λ) the characteristic polynomial, and the equation is called the characteristic
equation.
The basic equation is Ax=λx
𝟏 −𝟏 𝟎
Example Find the eigenvalues and eigenvectors of matrix A: −𝟏 𝟐 −𝟏
𝟎 −𝟏 𝟏
Cayley-Hamilton
𝟏
Example 1: use Cayley-Hamilton to find 𝑨−𝟏 Where 𝑨 =
𝟑
𝟏−𝝀
𝟐
𝑨 − 𝝀𝑰 = 𝟎 ⟹ 𝒅𝒆𝒕
=𝟎 ⟹
𝟑
𝟒−𝝀
𝑻𝒉𝒆 𝒄𝒉𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒊𝒔𝒕𝒊𝒄 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝑷 𝑨 = 𝟏 − 𝝀
𝑷 𝑨 =
𝝀𝟐
− 𝟓𝝀 − 𝟐 = 𝟎 ⟹
𝝀𝟐
− 𝟓𝝀 = 𝟐
𝟐
𝟒
𝟒−𝝀 −𝟔=𝟎 ⟹
𝑩𝒚 𝑪𝒂𝒚𝒍𝒆𝒚−𝑯𝒂𝒎𝒊𝒍𝒕𝒐𝒏 𝒕𝒉𝒆𝒐𝒓𝒆𝒎
𝟐 𝑰 = 𝑨𝟐 − 𝟓𝑨
Find 𝑨−𝟏
𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒃𝒚 𝑨−𝟏
𝑨−𝟏
𝟐𝑨−𝟏 𝑰 = 𝑨𝟐 𝑨−𝟏 − 𝟓𝑨𝑨−𝟏
𝟏
=
𝟐
𝟏
𝟑
𝟐
𝟏
−𝟓
𝟒
𝟎
𝟎
𝟏
𝑨−𝟏 =
𝟐𝑨−𝟏 = 𝑨 − 𝟓𝑰
𝑨−𝟏
𝟏 −𝟒
𝟐 𝟑
𝟐
−𝟏
𝟏
=
𝟐
𝟏
𝟑
𝑨−𝟏 =
𝟐
𝟓
−
𝟒
𝟎
𝟏
(𝑨 − 𝟓𝑰)
𝟐
𝟎
𝟓
Example 1: use Cayley-Hamilton to find and 𝑨𝟑 Where 𝑨 =
𝑷 𝑨 =
𝝀𝟐
− 𝟓𝝀 − 𝟐 = 𝟎
𝝀𝟐
= 𝟓𝝀 + 𝟐
𝟏 𝟐
𝟑 𝟒
𝑩𝒚 𝑪𝒂𝒚𝒍𝒆𝒚−𝑯𝒂𝒎𝒊𝒍𝒕𝒐𝒏 𝒕𝒉𝒆𝒐𝒓𝒆𝒎
𝑨𝟐 = 𝟓𝑨 + 𝟐 𝑰
𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒃𝒚 𝑨
𝑨𝟑 = 𝟓𝑨𝟐 + 𝟐 𝑨
𝑺𝒊𝒎𝒑𝒍𝒊𝒇𝒚 𝒕𝒉𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏
𝑩𝒖𝒕 𝑨𝟐 =𝟓𝑨+𝟐 𝑰
𝒕𝒉𝒆𝒏 𝑨𝟑 = 𝟓 𝟓𝑨 + 𝟐 𝑰 + 𝟐 𝑨
𝑨𝟑 = 𝟐𝟕 𝑨 + 𝟏𝟎 𝑰 ⇒ 𝑨𝟑 = 𝟐𝟕
𝟏
𝟑
𝟐
𝟏 𝟎
− 𝟏𝟎
𝟒
𝟎 𝟏
Example 1: Use Cayley-Hamilton to find 𝑪𝟐 𝒂𝒏𝒅 𝑪𝟖 Where 𝑪 =
𝟏
−𝝀
𝟐
𝑪 − 𝝀𝑰 = 𝟎 ⟹ 𝒅𝒆𝒕
𝟏
−
𝟐
⟹ 𝝀𝟐 = 𝝀
⟹ 𝑪𝟐 = 𝑪
−
𝟏
𝟐
𝟏
−𝝀
𝟐
=𝟎 ⟹
𝟏
−𝝀
𝟐
𝟏/𝟐 −𝟏/𝟐
−𝟏/𝟐 𝟏/𝟐
𝟏
𝟏
− 𝝀 − = 𝟎 ⟹ 𝝀𝟐 − 𝝀 = 𝟎
𝟐
𝟒
⟹ 𝑪𝟖 = 𝑪 ⟹ 𝑪𝟐 = 𝑪𝟖 = 𝑪𝒏 = 𝑪 =
𝟏/𝟐 −𝟏/𝟐
−𝟏/𝟐 𝟏/𝟐