Revision: Matrices Maths and Stats HSC Science (General) 12th Standard Board Exam Maharashtra State Board

Let \[A = [a_{ij}]\] be an \[m \times n\] matrix and \[B = [b_{jk}]\] be an \[n \times p\] matrix.

Then the product C = AB is an \[m \times p\] matrix \[C = [c_{ik}]\], where each entry \[c_{ik}\] is given by:

Consistent Solution: A system is consistent if it has at least one solution.

Inconsistent Solution: A system is inconsistent if it has no solution.

Two matrices are equivalent if one can be obtained from the other by a finite number of elementary operations

• Denoted by: A ∼ B

If A = [aij], then the negative of A, denoted by −A, is the matrix obtained by replacing each element a_ij by −a_ij

−A = [−a_ij]

• Order of −A = order of A

Minor

Delete ith row and jth column: M_ij​

Cofactor of a_ij

A_ij = (−1)^i+j × (minor of a_ij)

Sign pattern:

\[\begin{bmatrix}
+ & - & + \\
- & + & - \\
+ & - & +
\end{bmatrix}\]

Adjoint of A = transpose of the cofactor matrix \[\mathrm{adj}A=\left[A_{ij}\right]^T\]

• \[\mathrm{adj}(kA)=k^{n-1}\mathrm{adj}(A)\]

• A(adj A) = (adj A)A = ∣A∣I

• ∣adjA∣ = ∣A∣ⁿ⁻¹(for an n×n non-singular matrix)

• adj(AB) = (adjB)(adjA)

\[A=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}\]

\[A^{-1}=\frac{1}{ad-bc}
\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}\] if ad − bc ≠ 0

\[A^{-1}=\frac{1}{|A|}(\operatorname{adj}A)\], if ∣A∣ ≠ 0

Properties:

• \[(AB)^{-1}=B^{-1}A^{-1}\]

• \[(A^{-1})^{-1}=A\]

• \[(A^{\prime})^{-1}=(A^{-1})^{\prime}\]

• If inverse exists, it is unique.

• Matrix multiplication is row-by-column, not term-wise.

• Product AB exists only if columns of A = rows of B.

• If A is \[m \times n\] and B is \[n \times p\], then AB is \[m \times p\].

• In general, \[AB \neq BA\], and sometimes one product may not even be defined.

• Matrix multiplication is associative and distributive over addition.

• Identity matrix acts as a multiplicative identity: AI = IA = A.

• Zero matrix absorbs multiplication: AO = OA = O.

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

• Aⁿ is defined only when A is a square matrix.

• A^mAⁿ = A^m+n

• $I$

1. Write AX = B

2. Apply row operations on A
   (Same operations on B)

3. Reduce A to triangular/identity form

4. Solve equations

Comparable Matrices

• Two matrices are said to be comparable if they have the same order
  (same number of rows and columns).

Equal Matrices

Two matrices A= [a_ij] and B=[b_ij] are equal if:

1. They are comparable (same order), and

2. Their corresponding elements are equal.

Matrix Form: AX = B

Condition:

• A must be square

• ∣A∣ ≠ 0 (Non-singular)

Formula:

\[X=A^{-1}B\]​