Revision: 12th Std >> Matrices MAH-MHT CET (PCM/PCB) Matrices

If A and B are non-singular square matrices of the same order such that AB = BA = I (where I is the identity matrix of the same order as A and B), then A and B are called inverses of each other.

We write A⁻¹ = B and B⁻¹ = A.

i.e. AA⁻¹ = A⁻¹A = I.

• If |A| ≠ 0, then A⁻¹ exists.
• If the inverse of a square matrix exists, then it is unique. A matrix can not have more than one distinct inverse.

The adjoint of A is defined as the transpose (i.e. interchange rows and columns) of the cofactor matrix, and it is denoted by adj (A).

Let A = [a_ij] be a square matrix of order n. Then, the cofactor C_ij (or A_ij) of a_ij in A is (−1)^i+j times M_ij, where M_ij is the minor of a_ij in A.

∴ C_ij = (−1)^i+j M_ij

Let A = [a_ij] be a square matrix of order n. Then, the minor M_ij of a_ij in A is the determinant obtained by deleting the i^th row and the j^th column in which element a_ij lies. It is denoted by M_ij​ of A.

The adjoint of A is defined as the transpose (i.e. interchange rows and columns) of the cofactor matrix, and it is denoted by adj (A).

If A and B are non-singular square matrices of the same order such that AB = BA = I (where I is the identity matrix of the same order as A and B), then A and B are called inverses of each other.

We write A⁻¹ = B and B⁻¹ = A.

i.e. AA⁻¹ = A⁻¹A = I.

• If |A| ≠ 0, then A⁻¹ exists.
• If the inverse of a square matrix exists, then it is unique. A matrix can not have more than one distinct inverse.

The transpose of a matrix is obtained by interchanging its rows and columns.

• If a matrix is A, its transpose is denoted by A^T

• If A is of order m × n, then
  A^T is of order n × m

• First row of A becomes first column of A^T, and so on.

A determinant is a single real number associated with a square matrix only.

• Denoted by det ⁡A or ∣A∣ or Δ 

A square matrix \[A = [a_{ij}]_{n \times n}\] is called symmetric if

i.e., \[a_{ij} = a_{ji}\] for all i and j.

A square matrix \[A = [a_{ij}]_{n \times n}\] is called skew-symmetric if \[A^T = -A\]

i.e.,\[a_{ij} = -a_{ji}\] for all i and j.

By Adjoint Method: \[A^{-1}=\frac{\mathrm{adj}A}{|A|}\]

By Using Algebraic Equation: A matrix A and an algebraic equation in matrix A is in the form of A² + bA + C = O.

\[A^{-1}=\frac{1}{C}\left[-aA-bI\right]\]

By Adjoint Method: \[A^{-1}=\frac{\mathrm{adj}A}{|A|}\]

By Using Algebraic Equation: A matrix A and an algebraic equation in matrix A is in the form of A² + bA + C = O.

\[A^{-1}=\frac{1}{C}\left[-aA-bI\right]\]

Order 1 (1×1 matrix):

∣A∣ = a

Order 2 (2×2 matrix):

∣A∣ = ad − bc

Order 3 (3×3 matrix):

\[A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\]

\[|A|=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{21}a_{32}-a_{31}a_{22})\]

• If |A| = 0
  A matrix is called a Singular Matrix
• If |A| ≠ 0
  Matrix is called a Non-Singular Matrix

If A and B are symmetric matrices.

∴ A’ = A and B’ = B

(AB – BA) = (AB)’ – (BA)’   ...[∵ (X – Y) = X’ – Y’]

= B’A’ – A’B’   ...[∵ (XY) = Y’X’]

= BA – AB   ...[∵ B’ = B, A’ = A]

= –(AB – BA)

∴ AB – BA is a skew symmetric matrix.

Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Proof: Let A be a square matrix, then we can write

\[\mathrm{A=\frac{1}{2}(A+A^{\prime})+\frac{1}{2}(A-A^{\prime})}\]

From Theorem 1, we know that (A + A′) is a symmetric matrix and (A − A′) is a skew-symmetric matrix.

Multiplying by \[\frac{1}{2}\] does not change these properties.

Since for any matrix A, (kA)′ = kA′, it follows that \[\frac{1}{2}(\mathrm{A}+\mathrm{A}^{\prime})\] is symmetric matrix and \[\frac{1}{2}(\mathrm{A}-\mathrm{A}^{\prime})\] is skew symmetric matrix.

Thus, any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix. 

Theorem 1: For any square matrix A with real number entries, A + A′ is a symmetric matrix and A − A′ is a skew-symmetric matrix.

Proof:

Part 1: Symmetric Matrix
Let B = A + A′, then

Take transpose on both sides:
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= A + A′ (as A + B = B + A)
= B

Therefore, B = A + A′ is a symmetric matrix

Part 2: Skew-Symmetric Matrix

Now let
C = A − A′

C′ = (A − A′)′ = A′ − (A′)′ (Why?)
= A′ − A (Why?)
= −(A − A′) = −C

Therefore
C = A − A′ is a skew-symmetric matrix.

1. adj (AB) = (adj B) (adj A)
2. (adj A)A = A (adj A) = |A| Iₙ
3. (a) |adj A| = |A|ⁿ⁻¹, if |A| ≠ 0
   (b) |adj A| = 0, if |A| = 0
4. If |A| = 0, then (adj A) A = A (adj A) = O
5. adj (Aᵐ) = (adj A)ᵐ, m ∈ N
6. adj (kA) = kⁿ⁻¹ (adj A), k ∈ R
7. adj (adj A) = |A|ⁿ⁻² A, A is non-singular matrix
8. adj (adj A) = |A|ⁿ⁻² A, A is non-singular matrix

• Minor \[M_{ij}\]: determinant of the matrix obtained by deleting row i and column j.

• Cofactor \[C_{ij}\]: \[C_{ij} = (-1)^{i+j}M_{ij}\].

• Determinant expansion along row i: \[|A| = \sum_{j=1}^{n} a_{ij}C_{ij}\].

• Determinant expansion along column j: \[|A| = \sum_{i=1}^{n} a_{ij}C_{ij}\].

• Determinant value is the same for any choice of row or column for expansion.

• Mixed row/column property: \[\sum_{j=1}^{n} a_{ij}C_{kj} = 0\] for \[i \neq k\].

1. adj (AB) = (adj B) (adj A)
2. (adj A)A = A (adj A) = |A| Iₙ
3. (a) |adj A| = |A|ⁿ⁻¹, if |A| ≠ 0
   (b) |adj A| = 0, if |A| = 0
4. If |A| = 0, then (adj A) A = A (adj A) = O
5. adj (Aᵐ) = (adj A)ᵐ, m ∈ N
6. adj (kA) = kⁿ⁻¹ (adj A), k ∈ R
7. adj (adj A) = |A|ⁿ⁻² A, A is non-singular matrix
8. adj (adj A) = |A|ⁿ⁻² A, A is non-singular matrix

• Transpose = interchange rows and columns.

• If A is \[m \times n\], then A' is \[n \times m\].

• Standard notation: A' or \[A^T\].

• Key properties: (A')' = A, (kA)' = kA', (A + B)' = A' + B', (AB)' = B'A'.

• A square matrix is symmetric if \[A^T = A\].

• A square matrix is skew-symmetric if \[A^T = -A\].

• In a skew-symmetric matrix, all diagonal elements are zero.

• For any square matrix A:

  • \[A + A^T\] is symmetric.

  • \[A - A^T\] is skew-symmetric.

• Any square matrix A can be written as

• The decomposition into symmetric and skew-symmetric parts is unique.

Method of Inversion

• Given: AX = B
• Multiply both sides by A⁻¹

Result:

• X = A⁻¹B

Key Steps:

• Pre-multiply by A⁻¹
• Use property: A⁻¹A = I
• Final solution: X = A⁻¹B

Method of Reduction

• No need to find A⁻¹
• Apply elementary row operations to the matrix. Process:

• Convert the matrix into upper triangular form
• System reduces to:
  • b₁₁x + b₁₂y + b₁₃z = b₁′
  • b₂₂y + b₂₃z = b₂′
  • b₃₃z = b₃′

Final Step:

• Solve using back substitution:
  • First find z
  • Then y
  • Then x