Notes On Inverse of a Matrix - CBSE Class 12 Maths
The inverse of a matrix by using the elementary operations. If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A, and it is denoted by A-1. we have a square matrix, A, B of the order, n by n. A = $\left[\begin{array}{cccc}{\text{a}}_{\text{11}}& {\text{a}}_{\text{12}}& \cdots & {\text{a}}_{\text{1n}}\\ {\text{a}}_{\text{21}}& {\text{a}}_{\text{22}}& \text{⋯}& {\text{a}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{a}}_{\text{n1}}& {\text{a}}_{\text{n2}}& \text{⋯}& {\text{a}}_{\text{nn}}\end{array}\right]$  B =   AB = $\left[\begin{array}{cccc}{\text{a}}_{\text{11}}& {\text{a}}_{\text{12}}& \cdots & {\text{a}}_{\text{1n}}\\ {\text{a}}_{\text{21}}& {\text{a}}_{\text{22}}& \text{⋯}& {\text{a}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{a}}_{\text{n1}}& {\text{a}}_{\text{n2}}& \text{⋯}& {\text{a}}_{\text{nn}}\end{array}\right]$  ×  = I      Here, we can say that matrix A is invertible. The inverse of a matrix A does not mean that it is one divided by matrix A. i.e. A-1 ≠ 1/A A rectangular matrix does not possess inverse matrix, since for products AB and BA to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order. If B is the inverse of A, then A is also the inverse of B. Theorem: If the inverse of a matrix exists, it is unique. Proof : Let A be a square matrix of the order, n by n. A = [aij]nxn Let X and Y be inverse matrices of A. AX = XA = I AY = YA = I Now X = XI     = X(AY) ['.' AY = I]     = (XA)Y  ['.' Matrix multiplication is associative]     = IY  ['.' XA = I]  ⇒ X = Y Theorem: If A and B are invertible matrices of the same order, then (AB)-1 = B-1A-1. Proof : LHS: (AB)-1  (AB)(AB)-1 = I (The product of the matrices, AB and AB inverse, is equal to the identity matrix.) ⇒ A-1(AB)(AB)-1 = A-1I (Multiply with A-1 on both the sides) ⇒ (A-1A)B(AB)-1 = A-1  (Since matrix multiplication is associative) ⇒ IB(AB)-1 = A-1 ['.' A-1A = I] ⇒ B(AB)-1 = A-1 ⇒ B-1B(AB)-1 = B-1A-1 ⇒ I(AB)-1 = B-1A-1 ['.' B-1B = I] ⇒ (AB)-1 = B-1A-1 Inverse of a matrix by elementary operations Conder a marix A A = IA AA-1 = I The transformation is done by using either row operations or column operations. A = IA, A = IA Ex: X = $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-4}& \text{6}\end{array}\right]$ We can express this matrix as X = IX =  X        R2 → R2 + 4R1    =   X    R2 → 1/18 R2     =   X  R1 → R1 - 3R2    $\left[\begin{array}{cc}\text{1}& \text{0}\\ \text{0}& \text{1}\end{array}\right]$= $\left[\begin{array}{cc}\text{1/3}& \text{-1/6}\\ \text{2/9}& \text{1/18}\end{array}\right]$ X    the inverse of matrix X is X-1 = $\left[\begin{array}{cc}\text{1/3}& \text{-1/6}\\ \text{2/9}& \text{1/18}\end{array}\right]$      Now we will obtain the inverse of the same matrix, X, using column operations. X = XI $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-4}& \text{6}\end{array}\right]$   = X $\left[\begin{array}{cc}\text{1}& \text{0}\\ \text{0}& \text{1}\end{array}\right]$              Now we will perform column operations until the matrix on the left hand side transforms into an identity matrix. C2 → C2 - 3C1     = X      C2 → 1/18 C2      = X  C1 → C1 + 4C2     = X  X-1 = Note: While performing the row and column operations, if all the elements in a row or column are obtained as zero, then the matrix has no inverse.

#### Summary

The inverse of a matrix by using the elementary operations. If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A, and it is denoted by A-1. we have a square matrix, A, B of the order, n by n. A = $\left[\begin{array}{cccc}{\text{a}}_{\text{11}}& {\text{a}}_{\text{12}}& \cdots & {\text{a}}_{\text{1n}}\\ {\text{a}}_{\text{21}}& {\text{a}}_{\text{22}}& \text{⋯}& {\text{a}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{a}}_{\text{n1}}& {\text{a}}_{\text{n2}}& \text{⋯}& {\text{a}}_{\text{nn}}\end{array}\right]$  B =   AB = $\left[\begin{array}{cccc}{\text{a}}_{\text{11}}& {\text{a}}_{\text{12}}& \cdots & {\text{a}}_{\text{1n}}\\ {\text{a}}_{\text{21}}& {\text{a}}_{\text{22}}& \text{⋯}& {\text{a}}_{\text{2n}}\\ ⋮& \text{⋮}& \text{⋮}& \text{⋮}\\ {\text{a}}_{\text{n1}}& {\text{a}}_{\text{n2}}& \text{⋯}& {\text{a}}_{\text{nn}}\end{array}\right]$  ×  = I      Here, we can say that matrix A is invertible. The inverse of a matrix A does not mean that it is one divided by matrix A. i.e. A-1 ≠ 1/A A rectangular matrix does not possess inverse matrix, since for products AB and BA to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order. If B is the inverse of A, then A is also the inverse of B. Theorem: If the inverse of a matrix exists, it is unique. Proof : Let A be a square matrix of the order, n by n. A = [aij]nxn Let X and Y be inverse matrices of A. AX = XA = I AY = YA = I Now X = XI     = X(AY) ['.' AY = I]     = (XA)Y  ['.' Matrix multiplication is associative]     = IY  ['.' XA = I]  ⇒ X = Y Theorem: If A and B are invertible matrices of the same order, then (AB)-1 = B-1A-1. Proof : LHS: (AB)-1  (AB)(AB)-1 = I (The product of the matrices, AB and AB inverse, is equal to the identity matrix.) ⇒ A-1(AB)(AB)-1 = A-1I (Multiply with A-1 on both the sides) ⇒ (A-1A)B(AB)-1 = A-1  (Since matrix multiplication is associative) ⇒ IB(AB)-1 = A-1 ['.' A-1A = I] ⇒ B(AB)-1 = A-1 ⇒ B-1B(AB)-1 = B-1A-1 ⇒ I(AB)-1 = B-1A-1 ['.' B-1B = I] ⇒ (AB)-1 = B-1A-1 Inverse of a matrix by elementary operations Conder a marix A A = IA AA-1 = I The transformation is done by using either row operations or column operations. A = IA, A = IA Ex: X = $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-4}& \text{6}\end{array}\right]$ We can express this matrix as X = IX =  X        R2 → R2 + 4R1    =   X    R2 → 1/18 R2     =   X  R1 → R1 - 3R2    $\left[\begin{array}{cc}\text{1}& \text{0}\\ \text{0}& \text{1}\end{array}\right]$= $\left[\begin{array}{cc}\text{1/3}& \text{-1/6}\\ \text{2/9}& \text{1/18}\end{array}\right]$ X    the inverse of matrix X is X-1 = $\left[\begin{array}{cc}\text{1/3}& \text{-1/6}\\ \text{2/9}& \text{1/18}\end{array}\right]$      Now we will obtain the inverse of the same matrix, X, using column operations. X = XI $\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-4}& \text{6}\end{array}\right]$   = X $\left[\begin{array}{cc}\text{1}& \text{0}\\ \text{0}& \text{1}\end{array}\right]$              Now we will perform column operations until the matrix on the left hand side transforms into an identity matrix. C2 → C2 - 3C1     = X      C2 → 1/18 C2      = X  C1 → C1 + 4C2     = X  X-1 = Note: While performing the row and column operations, if all the elements in a row or column are obtained as zero, then the matrix has no inverse.

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