Definition: A system of linear equation in unknowns x_{1,} x_{2}, .... , x_{n} is of the form
If b’s are all zero this system is called homogeneous and if at least one of b_{i}'s is nonzero then the above system is called nonhomogeneous.
Case I. NonHomogeneous linear equation (Two unknowns):
consider the system of equations
The following three cases arise
Let b_{2} ≠ 0. Then the system (1) is equivalent to
It is obvious that if pair (x_{0}, y_{0}) is a solution of the system (1). It is also a solution of system (2) and viceversa.
Multiplying the second equation of (2) by b, and subtracting it from the first equation of (2), we get
Now replacing the first equation of the system (2) by the equation (3), the system (2) is equivalent to the system of equations
Now consider the following subcases:
the first equation of the system (4) gives
Substituting this value of x into the second equation of system (4), we get
For convenience, we write
Any set of values of x_{1}, x_{2}, ...., x_{n} (real numbers or complex numbers if the coefficients are complex number) which simultaneously satisfy each of these equations is called solution of the system (1).
A system of equations is called consistent if it has at least one solution, inconsistent if it does not have any solution.
If the system of equations has a unique solution it is called determinate, it is said to be indeterminate if it has more than one solution.
➤ Nonhomogeneous Equations
Here Δ_{x} and Δ_{y} are obtained by replacing the elements in the first and second columns in A by the elements c_{1}, c_{2}. Thus (5) and (6) can be written as
This method of solving simultaneous equations is known as Cramer’s rule. Then if a_{1}b_{2}  a_{2}b_{1} ≠ 0 then the system (4) or system (1) has the unique solution given by (7). Hence the equations are consistent and determinate.
When Δ = a_{1}b_{2}  a_{2}b_{1} = 0.
Then the system (4) becomes
Obviously this system (8) has no solution if c_{1}b_{2}  c_{2}b_{1} = Δ_{x} ≠0. Therefore the equations are inconsistent.
But if c_{1}b_{2}  c_{2}b_{1} = Δ_{x} =0, then any pair of number (x, y), where
is a solution of system (8).
Hence in this case, the equations are consistent and indeterminate. We from the above discussion, conclude that
Case II. Nonhomogeneous linear equations in three unknowns.
Consider the system of three nonhomogeneous linear equations in three unknowns x, y, z
Introduce the following notations
For testing the consistency of the given system of equations (1), we give the following rule.
➤ Solution of simultaneous Homogeneous linear equations
Two unknowns: Consider the system of linear equations
The system (1) always has the trivial solution x = 0, y = 0 so it is always consistent.
The system (1) has a nontrivial solution iff Δ = 0 and in this case the equations have infinite number of solutions. If Δ ≠ 0 then the system (1) has unique solution x = 0, y = 0 similarly, we can discuss the solution of a system of three homogeneous linear equations in three unknowns x, y, z for such a system if Δ ≠ 0, then it has the unique solution x = 0, y = 0, z = 0.
If A = 0 then the system has an infinite number solutions.
Thus such a system of equations is always consistent.
Three equations in two unknowns consider the system of equations
The system (1) will be consistent if the values of x, y obtained from any two equations also satisfy the third equation i.e. if
➤ Solution of Simultaneous Linear NonHomogeneous Equations
Let us consider the system of n simultaneous equations having n unknowns x_{1}, x_{2}, x_{3} .... x_{n}; as given below:
(No of unknowns = No of equations) or in compact form, the above system can be written as
where i = 1, 2,..... n.
In matrix form, equation (1) can be put as
⇒ AX = B
where
is known as the coefficient matrix.
X = transposed matrix of [x_{1}, x_{2},x_{3}, ... x_{n} ]
and B = transposed matrix of [b_{1}, b_{2},b_{3}, ... b_{n} ]
when A is a nonsingular matrix, then A^{1} exists and hence A^{1} A = I, where I is the identity matrix 1st Method. From above, we have AX = B
A^{1} AX = A^{1}B ⇒ I X = A^{1} B ⇒ X = A^{1}B
From here, one can obtain x_{1}, x_{2}, ... x_{n}. This solution is unique.
When the system of equations has one or more solutions, the equations are said to be consistent, otherwise the equations can said to be inconsistent.
2nd Method: To solve the equation AX = B, we perform Erow operations on the both A and B so that the matrix A is reduced to the triangular form. Let A be reduced to A_{1} and B to B_{1}; Then AX = B give A_{1}X_{1} = B_{1} which when solved gives x_{1},x_{2}, ..., x_{n}.
Example. Solve by matrix method, the system x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6.
The given system in matrix from can be put as
⇒ AX = B
⇒ X = A^{1}B
To find A^{1}, we compute Adj A and I A I as follows:
Also
= 1 (18  12) 1 (9  3) + 1 (4  2) = 2
Consistent: A system of equations is said to be consistent if they have one or more solution,
Inconsistent: When a particular system of equations has no solution at all, it is said to be inconsistent e.g., x + 2y = 4, 3x + 6x = 5.
Consistency of a set of m equations (nonhomogeneous) in n unknowns
⇒ AX = B ..(1)
Then we obtain the augmented matrix:
For the solution of AX = B, proceed as follows.
Here we have m nonhomogeneous equations in n unknowns, then matrix A is of order m x n. Write the augmented matrix [A I B] and reduce it to the Echelon form by applying elementary row operations only. Thus we shall be able to find the rank of augmented matrix [A I B]. Also we can find the rank coefficient matrix A by deleting the last column of the Echelon form [A I B]. Note that deleting the last column of [A I B] we get the Echelon form of coefficient matrix A.
Now there are two possibilities.
From the Echelon form of [A I B], we write the equivalent system of equations. If r<m, then in the process of reducing [A I B] to Echelon form mr rows are reduced to zero rows and so mr equations will be eliminated. Thus here we obtain an equivalent system of r equations only. From these r equations we shall be able to express the values of r unknowns in terms of the remaining nr unknowns. We can give any arbitrary chosen values to these nr unknowns. Now there exist the following cases.
I. If r=n, then nr=0. In this case there will be an unique solution of the equations as no variable will be given an arbitrary value.
II. If r<n, then nr >0. In this case arbitrary value will be given to the nr variables. Thus in this case there will be an infinite number of solutions.
III. If m<n, then r<m<n, then nr>0. In this case also arbitrary values will be given to the n  r variables. Thus in this case there will be an infinite number of solutions.
Example. Test for consistency and hence solve x + y + z = 3, 3x + y  2z = 2, 2x + 4y + 7z = 7.
We have
and
Now
or
But rank of the R.H.S matrix is 3 as
so ρ([AB]) = 3
Also
or
Now
∴ ρ(A) = 2
ρ(A) ≠ ρ([AB]) and hence the given system of equations is inconsistent.
where the number of equations (m here) may be different from the number of unknown. The above system in matrix form takes the form AX = O ...(1)
where
Here A and augmented matrix [A I B] i.e [A I 0] are the same so that ρ(A) = rank of the augmented matrix. Hence the system of homogenous equations is always consistent ⇒ x_{1} = x_{2} = ... = x_{n} = 0 is always a solution if det A ≠ 0 which is called a trivial solution or zero solution. But we are interested in nontrivial solution.
Case (i) Let ρ(A) = r = n(≤ m)
In this case the number of variables to be assigned arbitrary values in n  r = 0. Hence ∃ a unique solution X = 0 i.e x_{1} = x_{2} = . . = x_{n} = 0.
Case (ii) ρ(A) = r < n(≤ m)
Here (n  r) variables can be selected and assigned arbitrary values and hence ∃ infinite number of solutions.
When m < n, ρ(A) = r ≤ m < n, (n  r) Variables can be selected and assigned arbitrary values. Hence when the number of equations is less then the number of unknowns, the equations will always have an infinite number of solutions.
Example. Solve the system 2x  y + 3z = 0, 3x + 2y + z = 0 and x  4y + 5z = 0.
Here
⇒ ρ(A) = 2 < 3
(i.e rank is less than the number of variables)
Now, the matrix form of the system reduces to
⇒ x = z, y = z. Choosing z = k,
we get x = k, y = k, z = k.
This gives the general solution of the given system where k is any arbitrary constant.
Definitions
Let A = [a_{ij}]_{n*n} be a square matrix of order n, I is an unit matrix of order n and λ an indeterminate, then the matrix
is called the characteristic matrix of A and the determinant
is called the characteristic polynomial of A.
Also the equation IA  λII = 0, is called the characteristic equation of A. The roots of this equations are called the characteristic roots or characteristic value or Eigen roots or Eigen values of latent roots of the matrix.
The set of the Eigen values of the matrix A is called the spectrum of the matrix A If X is a characteristic root of a n * n matrix A, then the nonzero solution
of the equation AX = λX i.e. (A  λI) X = 0 is called the characteristic vector or Eigen vector of the matrix A corresponding to the characteristic root λ.
Chief Characteristics of Eigen Values
The sum of the elements of the principal diagonal of a matrix is called the trace of the matrix.
Diagonalisation of a matrix
Definition: Any matrix A is said to diagonalisable if A can be written as PDP^{1} where D is diagonal matrix.
In that case, f(A) = f(PDP ^{1}) = P f(D)P^{1} for any polynomial. This is useful as it’s easier to calculate f(D).
Theorem: λ is a characteristic root of a matrix A if and only if there exists a nonzero vector X such that AX = λX.
Theorem: The matrix A and its transpose have the same characteristic roots.
Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.
Hence if the matrices A and AT have the same characteristic polynomial, then they have the same eigenvalues.
So we show that the characteristic polynomial p_{A}(t) = det (A  tl) of A is the same as the characteristic polynomial of the transpose A^{T}.
We have
= det(A^{T}  tl^{T}) since l^{T} = I
= det((A  tl)^{T})
= det(A  tl) since det (B^{T}) = det(B) for any square matrix B
= P_{a}(t).
Therefore we obtain , and we conclude that the eigenvalues of A and A^{T} are the same.
Theorem: If A and P be square matrices and A~B and if P is invertible, then the matrices A and P^{1} AP have the same characteristic roots.
Theorem: If A and B be n rowed square matrices and if A be invertible, then the matrices A^{1} B and BA^{1} have the same characteristic roots.
Theorem: 0 is a characteristic root of a matrix if and only if the matrix is singular.
Theorem: If A and B be two square invertible matrices then AB and BA have, the same characteristics roots.
Theorem: The characteristic roots of a triangular matrix are just the diagonal elements of the matrix.
Theorem: All the characteristics roots of a Hermitian matrix are real.
Cor. : Characteristic roots of a real symmetric matrix are all real.
Cor. : Characteristic roots of skew Hermitian matrix is either zero or a pure imaginary number.
Theorem: Any two characteristic vectors corresponding to two distinct characteristic roots of a : (i) Hermitian, (ii) real symmetric, (iii) Unitary matrix are orthogonal.
Theorem: The characteristic vector corresponding to characteristic root X of matrix A is also a characteristic vector of every matrix f(A) where f(x) is any scalar polynomial and the corresponding root for f(A) is f(λ) and in general.
if and f_{2}(A) ≠ 0 then g(λ) is a characteristic root of g(A) = f_{1}(A) .{f_{2}(A)}^{1}_{.}
_{Theorem: The modulus of each characteristic root of a unitary matrix is unity.}
Theorem: Any system of characteristic vectors X_{1}, X_{2}, .... X_{k} respectively corresponding to a system of distinct characteristic roots λ1,λ2, .... λ_{k} of any square matrix A are linearly independent.
Theorem: A matrix is nilpotent iff 0 is the only eigenvalue.
(⇒)
Suppose the matrix A is nilpotent. Namely there exists k ∈ ℕ such that A^{k} = O. Let λ be an eigenvalue of A and let x be the eigenvector corresponding to the eigenvalue λ.
Then they satisfy the equality Ax = λx. Multiplying this equality by A on the left, we have
A^{2}x = λAx = λ^{2}x.
Repeatedly multiplying by A, we obtain that A^{k}x = λ^{k}x. (To prove this statement, use mathematical induction.)
Now since A^{k} = O, we get λ^{k}x = 0_{n}, ndimensional zero vector.
Since x is an eigenvector and hence nonzero by definition we obtain that λ^{k} = 0, and hence λ = 0.
(⇐)
Now we assume that all the eigenvalues of the matrix A are zero.
We prove that A is nilpotent.
There exists an invertible n * n matrix P such that P^{1} AP is an upper triangular matrix whose diagonal entries are eigenvalues of A.
(This is always possible. Study a triangularizable matrix or Jordan normal/canonical form.)
Hence we have
Then we have (P^{}^{1} AP)^{n} = O. This implies that P^{1} A^{n} P = O and thus A^{n} = POP^{1} = O.
Therefore the matrix A is nilpotent.
Theorem: The characteristic polynomial of an n x n matrix A has the form
p(t) = t^{n}  (trace A )t^{n}  1 + (intermediate terms) + (1)^{n}(detA)
Theorem: Every complex matrix A can be written as PTP1 where T is triangular matrix. Theorem: Let A be a unitary matrix. There is a unitary matrix P such that P*AP is diagonal. In particular all unitary and hence symmetric matrices are diagonalisable.
Note: Proof of above theorems is not required. You can directly use them as results.
Example. Determine the Eigen values and Eigen vectors of the matrix
.
= (5  λ)(2  λ)  4 =λ^{2}  7λ + 6
IA  λII = λ^{2}  7λ + 6 = 0
⇒ (λ  1 )(λ  6) = 0
∴ λ = 1, 6. Hence the Eigen values of matrix A are 1, 6.
Now the Eigen vector of matrix A corresponding to the Eigen vector of matrix A corresponding to the Eigen value λ is given by the nonzero solution of the equation (A  λI) X = 0
or
...(1)
When λ = 1, the corresponding Eigen vector is given by
or 4x_{1}+ 4x_{2} = 0 and x_{1} + x_{2} = 0
From which we get x, = x_{2}
or
may be taken as an Eigen vector of A corresponding to the Eigen value λ = 1
When λ = 6, substituting in (1) the corresponding Eigen vector is given by
From which we get, x_{1} = 4x^{2}
Thus may be taken as an Eigen vector of A corresponding to the Eigen value λ = 6.
Hence the Eigen vectors are
Every square matrix satisfies its own characteristic equation.
or
For a square matrix A of order n, if
IA  λII = (  1)^{n} (λ^{n} + a_{1}λ^{n1} + a_{2}λ^{n2} + .... + a_{n})
be the characteristic polynomial, then the matrix equation X^{n} + a_{1}X^{n1}+ a_{2}X^{n2}_{ }+_{ } .... + a_{n} I = 0 is satisfied by X = A i.e., A^{n} + a_{1}A^{n1} + a_{2}A^{n2} + .... + a_{n}I = 0.
Cor. Inverse of matrix by CayleyHamilton Theorem
IA  λII = a_{0}λ^{n} + a_{1}λ^{n1} + a_{2}λ^{n2} + .... + a_{n1}λ + a_{n}
is the characteristic polynomial of A, so we have
a_{0}A^{n1} + a_{1}A^{n1} + ... + a_{n1}A + a_{n} I = 0 ... (1)
Now premultiplying (1) by A^{1}, we get
Example. Verify Cayley Hamilton Theorem for the matrix Hence find A^{1}.
We have det (A  λI) = IA  λII
Now the characteristic equation of A is given byIA  λII =  λ^{3} + 6λ^{2}  7λ  2 = 0
⇒ λ3  6λ^{2} + 7λ + 2 = 0 ...(1)
By CayleyHamilton theorem, the matrix A must satisfy (1) i.e. A^{3}  6A^{2} + 7A + 2I = 0
Now
⇒ CayleyHamilton theorem is verified.
To find A^{1 }
we have 2I =  A^{3} + 6A^{2}  7A
Then premultiplying both side by A^{1}, we get
2A^{1} I =  A^{1} A. + 6A^{1} A^{2}  7A^{1} A
2A^{1} =  IA^{2} + 6IA  7I ( ∴ A^{}^{1} A = I and IA = A etc)
⇒ 2A^{1} = ( A^{2} + 6A  7I)
Theorem: If A is an n * n matrix with
then
Theorem: If A is an n x n upper (or lower) triangular matrix, the eigenvalues are the entries on its main diagonal.
Definition: Let A be an n * n matrix and let
(1) The number n_{1}, n_{2}, n_{3}, ..., n_{k} are the algebraic multiplicities of the eigenvalues r_{1,}r_{2}, r_{3}, ... , respectively.
(2) The geometric multiplicity of the eigenvalue r_{j} (j = 1, 2, 3 ...... k) is the dimension of the null space A  r_{j}ln (j = 1, 2, 3, ... , k).
for example
The above examples suggest the following theorem:
Theorem: Let A be an n x n matrix with eigenvalue λ. Then the geometric multiplicity of λ is less than or equal to the algebraic multiplicity of λ.
Objective: Solve
with an n x n constant coefficient matrix A.
Here, the unknown is the vector function
General Solution Formula in Matrix Exponential Form:
where C_{1},......C_{n} are arbitrary constants.
The solution of the initial value problem
is given by
Definition (Matrix Exponential): For a square matrix A,
Evaluation of Matrix Exponential in the Diagonalizable Case: Suppose that A is diagonalizable; that is, there are an invertible matrix P and a diagonal matrix such that A = PDP^{1}. In this case, we have
Example:
(a) Evaluate e^{tA}
_{(b) Find the general solutions of .}
_{(c) Solve the initial value problem }
_{The given matrix A is diagonalized: A = PDP}^{1}_{ with }
_{}
_{Part (a): We have }
_{e}^{ta}_{ = Pe}^{tD}_{ P}^{1}
_{}
Part (b): The general solutions to the given system are
where C_{1}, C_{2,} C_{3} are free parameters.
Part (c): The solution to the initial value problem is
Evaluation of Matrix Exponential Using Fundamental Matrix: In the case A is not diagonalizable, one approach to obtain matrix exponential is to use Jordan forms. Here, we use another approach. We have already learned how to solve the initial value problem
We shall compare the solution formula with to figure out what e^{tA} is. We know the general solutions of _{} are of the following structure:
where are n linearly independent particular solutions. The formula can be rewritten as
For the initial value problem is determined by the initial condition
Thus, the solution of the initial value problem is given by
Comparing this with , we obtain
In this method of evaluating e^{tA}, the matrix plays an essential role. Indeed, e^{tA} = M(t) M(0)^{1}.
Definition (Fundamental Matrix Solution): if are n linearly independent solutions of the n dimensional homogeneous linear system we call
a fundamental matrix solution of the system.
(Remark 1: The matrix function M(t) satisfies the equation M’(t) = AM(t). Moreover, M(t) is an invertible matrix for every t. These two properties characterize fundamental matrix solutions.)
(Remark 2: Given a linear system, fundamental matrix solutions are not unique. However, when we make any choice of a fundamental matrix solution M(t) and compute M(t)M(0)^{1} we always get the same result.)
Example 1: Evaluate e^{tA} for
We first solve We obtain
This gives a fundamental matrix solution:
The matrix exponential is
Example 2: Evaluate e^{tA} for
We first solve We obtain
This gives a fundamental matrix solution:
The matrix exponential is
Example 3: Evaluate e^{tA} for
 (Use Diagonalization)
Solving det(A  λI) = 0, we obtain the eigenvalues of A: λ_{1} = 7 + 4i, λ_{2} = 7  4i.
Eigen vectors for λ_{1} = 7 + 4i; are obtained by solving
Eigenvectors for λ_{2} = 7  4i: are complex conjugate of the vector in (*). The matrix A is now diagonalized: A = PDP^{1} with
we have
etA = Pe^{tD}_{ P}^{1} (Use fundamental solutions and complex exp functions)
A fundamental matrix solution can be obtained from the eigenvalues and eigenvectors:
The matrix exponential is (Use fundamental solutions and avoid complex exp functions)
A fundamental matrix solution can be obtained from the eigenvalues and eigenvectors :
The matrix exponential is
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1. What is a system of linear equations? 
2. What does it mean for a system of linear equations to be consistent? 
3. What does it mean for a system of linear equations to be inconsistent? 
4. What is the importance of eigenvalues and eigenvectors in linear algebra? 
5. How can eigenvalues and eigenvectors be computed? 

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