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Matrices - Invertible matrices and proof of the uniqueness of inverse

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of an Invertible Matrix: A square matrix AA of order mm is called invertible if there exists another square matrix BB of the same order such that AB=BA=IAB = BA = I, where II is the identity matrix of order mm. In this case, BB is called the inverse of AA and is denoted by A−1A^{-1}. Visually, the identity matrix II functions as a grid where the diagonal from top-left to bottom-right consists of 1s, representing no change in transformation.

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Uniqueness of Inverse: The inverse of a square matrix, if it exists, is unique. This means a matrix cannot have two different inverses. If we hypothesize two inverses BB and CC for matrix AA, we can mathematically demonstrate that B=CB = C, ensuring that every invertible matrix maps to exactly one inverse matrix.

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Necessity of Square Matrices: For a matrix to be invertible, it must be a square matrix (n×nn \times n). A rectangular matrix (m×nm \times n where m≠nm \neq n) cannot have an inverse because for the products ABAB and BABA to be defined and equal to the identity matrix, the dimensions must match perfectly. Visually, only a perfectly square grid can be mirrored across its diagonal to return to its original form through an inverse operation.

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Inverse of a Product (Reversal Law): If AA and BB are invertible matrices of the same order, then (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}. This is known as the reversal law. Imagine this as a sequence of transformations: if you apply transformation AA then BB, to undo the result, you must first undo BB (the last action) and then undo AA (the first action).

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Elementary Operations and Invertibility: A matrix AA is invertible if and only if it can be reduced to the identity matrix II using elementary row or column operations. If, while performing these operations, we obtain a row or column of all zeros, the matrix is singular and does not have an inverse. Visually, a row of zeros indicates that the matrix 'collapses' a dimension, making it impossible to reconstruct the original input.

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The Identity Matrix Property: The identity matrix II acts as the multiplicative identity such that AI=IA=AAI = IA = A. In a coordinate system, multiplying by the identity matrix is equivalent to a transformation that leaves every point in its original position, which is why A×A−1A \times A^{-1} results in II—the transformations effectively cancel each other out.

📐Formulae

AB=BA=IAB = BA = I

B=A−1B = A^{-1}

(A−1)−1=A(A^{-1})^{-1} = A

(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T

(kA)−1=1kA−1, where k≠0(kA)^{-1} = \frac{1}{k}A^{-1}, \text{ where } k \neq 0

💡Examples

Problem 1:

Prove that the inverse of a square matrix, if it exists, is unique.

Solution:

Let AA be a square matrix of order mm. Suppose BB and CC are two inverses of AA. \ Since BB is the inverse of AA, we have: \ AB=BA=IAB = BA = I ... (i) \ Since CC is also the inverse of AA, we have: \ AC=CA=IAC = CA = I ... (ii) \ Now, consider BB: \ B=BIB = BI (Property of Identity Matrix) \ B=B(AC)B = B(AC) (Substituting I=ACI = AC from (ii)) \ B=(BA)CB = (BA)C (Associative property of matrix multiplication) \ B=ICB = IC (Substituting BA=IBA = I from (i)) \ B=CB = C \ Hence, the inverse is unique.

Explanation:

This proof uses the associative property of matrix multiplication and the property of the identity matrix to show that any two assumed inverses must actually be the same matrix.

Problem 2:

If A=[2312]A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} and B=[1011]B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

Solution:

Step 1: Calculate ABAB. \ AB=[2312][1011]=[2(1)+3(1)2(0)+3(1)1(1)+2(1)1(0)+2(1)]=[5332]AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2(1)+3(1) & 2(0)+3(1) \\ 1(1)+2(1) & 1(0)+2(1) \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 3 & 2 \end{bmatrix} \ Step 2: Find (AB)−1(AB)^{-1}. \ det(AB)=(5)(2)−(3)(3)=1\text{det}(AB) = (5)(2) - (3)(3) = 1. \ (AB)−1=11[2−3−35]=[2−3−35](AB)^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix} \ Step 3: Find B−1B^{-1} and A−1A^{-1}. \ det(B)=1(1)−0(1)=1  ⟹  B−1=[10−11]\text{det}(B) = 1(1)-0(1) = 1 \implies B^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \ det(A)=2(2)−3(1)=1  ⟹  A−1=[2−3−12]\text{det}(A) = 2(2)-3(1) = 1 \implies A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \ Step 4: Calculate B−1A−1B^{-1}A^{-1}. \ B−1A−1=[10−11][2−3−12]=[1(2)+0(−1)1(−3)+0(2)−1(2)+1(−1)−1(−3)+1(2)]=[2−3−35]B^{-1}A^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1(2)+0(-1) & 1(-3)+0(2) \\ -1(2)+1(-1) & -1(-3)+1(2) \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix} \ Since Step 2 and Step 4 match, (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

Explanation:

This example verifies the reversal law for matrix inverses by calculating both sides of the equation independently using the 2×22 \times 2 inverse formula: M−1=1ad−bc[d−b−ca]M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.