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Question

For a square matrix A, which of the following properties hold?

1) (A -1 )-1 = A

2) \(\det \left( {{A^{ - 1}}} \right) = \frac{1}{{detA}}\)

3) (λA) -1 = λA -1 where λ is a scalar

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This question was previously asked in
NDA I 2018 GAT Previous Year Paper (22-Apr-2018)
The correct answer is

1 and 2 only

Understanding Matrix Properties for Square Matrices

This question asks about the properties of a square matrix A, specifically concerning its inverse and determinant. We need to analyze each given property to determine its validity.

Analyzing Property 1: Inverse of the Inverse

The first property states that \( \left( {{A^{ - 1}}} \right)^{ - 1} = A \). Let's consider what an inverse matrix means. A matrix \( B \) is the inverse of matrix \( A \) if \( AB = BA = I \), where \( I \) is the identity matrix. We denote \( B \) as \( A^{-1} \).

Now consider the inverse of \( A^{-1} \). Let \( C = (A^{-1})^{-1} \). By definition, \( C \) is the matrix such that \( (A^{-1})C = C(A^{-1}) = I \). We know that \( A(A^{-1}) = (A^{-1})A = I \). Comparing these equations, we can see that \( C \) must be equal to \( A \).

Therefore, the inverse of the inverse of a square matrix A is indeed the matrix A itself, provided A is invertible.

Property 1 holds true.

Analyzing Property 2: Determinant of the Inverse

The second property states that \( \det \left( {{A^{ - 1}}} \right) = \frac{1}{{\det A}} \). For this property to be meaningful, the matrix A must be invertible, which implies that \( \det(A) \neq 0 \).

We know a fundamental property of determinants related to matrix multiplication: \( \det(AB) = \det(A) \det(B) \). Using this property, we can consider the relationship between A and its inverse \( A^{-1} \).

We have \( A A^{-1} = I \). Taking the determinant of both sides:

\( \det(A A^{-1}) = \det(I) \)

Using the multiplication property of determinants on the left side, and knowing that the determinant of the identity matrix I is 1:

\( \det(A) \det(A^{-1}) = 1 \)

Assuming \( \det(A) \neq 0 \), we can divide both sides by \( \det(A) \):

\( \det(A^{-1}) = \frac{1}{\det(A)} \)

This derivation shows that the property is correct for an invertible square matrix A.

Property 2 holds true.

Analyzing Property 3: Inverse of a Scalar Multiple

The third property states that \( (\lambda A)^{-1} = \lambda A^{-1} \), where \( \lambda \) is a scalar. Let's consider the properties of matrix inversion and scalar multiplication.

For a scalar \( \lambda \neq 0 \) and an invertible matrix A, the inverse of \( \lambda A \) is given by \( (\lambda A)^{-1} = \frac{1}{\lambda} A^{-1} \). We can verify this by multiplying \( (\lambda A) \) by \( \frac{1}{\lambda} A^{-1} \):

\( (\lambda A) \left( \frac{1}{\lambda} A^{-1} \right) = \lambda \left( \frac{1}{\lambda} \right) (A A^{-1}) = 1 \cdot I = I \)

And similarly, \( \left( \frac{1}{\lambda} A^{-1} \right) (\lambda A) = \left( \frac{1}{\lambda} \right) \lambda (A^{-1} A) = 1 \cdot I = I \).

This confirms that the correct formula for the inverse of \( \lambda A \) is \( \frac{1}{\lambda} A^{-1} \). The property stated, \( (\lambda A)^{-1} = \lambda A^{-1} \), would only be true if \( \lambda = \frac{1}{\lambda} \), which implies \( \lambda^2 = 1 \), so \( \lambda = \pm 1 \). Since the property is stated for *a* scalar \( \lambda \) without restriction to \( \pm 1 \), it is not generally true.

For example, if \( A = \begin{pmatrix} 2 \end{pmatrix} \) and \( \lambda = 2 \), then \( A^{-1} = \begin{pmatrix} 1/2 \end{pmatrix} \). \( \lambda A = 2 \begin{pmatrix} 2 \end{pmatrix} = \begin{pmatrix} 4 \end{pmatrix} \). So \( (\lambda A)^{-1} = \begin{pmatrix} 1/4 \end{pmatrix} \). According to property 3, \( (\lambda A)^{-1} = \lambda A^{-1} = 2 \begin{pmatrix} 1/2 \end{pmatrix} = \begin{pmatrix} 1 \end{pmatrix} \). Clearly, \( \begin{pmatrix} 1/4 \end{pmatrix} \neq \begin{pmatrix} 1 \end{pmatrix} \). This simple example shows property 3 is false.

Property 3 does not hold true for all scalars \( \lambda \).

Conclusion on Matrix Properties

Based on our analysis:

  • Property 1: \( (A^{-1})^{-1} = A \) is true.
  • Property 2: \( \det(A^{-1}) = \frac{1}{\det(A)} \) is true.
  • Property 3: \( (\lambda A)^{-1} = \lambda A^{-1} \) is false for a general scalar \( \lambda \).

Therefore, only properties 1 and 2 hold for a square matrix A (assuming it is invertible for properties 1 and 2 to be defined).

Property Statement Validity
1 \( (A^{-1})^{-1} = A \) True
2 \( \det(A^{-1}) = \frac{1}{{\det A}} \) True
3 \( (\lambda A)^{-1} = \lambda A^{-1} \) False (Correct is \( \frac{1}{\lambda} A^{-1} \))

Revision Table: Key Matrix Inverse and Determinant Properties

Concept Property Condition(s)
Inverse of Inverse \( (A^{-1})^{-1} = A \) A is invertible
Determinant of Inverse \( \det(A^{-1}) = \frac{1}{{\det A}} \) A is invertible
Inverse of Scalar Multiple \( (\lambda A)^{-1} = \frac{1}{\lambda} A^{-1} \) A is invertible, \( \lambda \neq 0 \)
Determinant of Scalar Multiple \( \det(\lambda A) = \lambda^n \det(A) \) A is \( n \times n \) square matrix
Determinant of Product \( \det(AB) = \det(A)\det(B) \) A, B are square matrices of same size

Additional Information: Understanding Invertible Matrices

A square matrix A is said to be invertible or non-singular if there exists a square matrix \( A^{-1} \) of the same size such that \( AA^{-1} = A^{-1}A = I \), where I is the identity matrix. If a matrix is not invertible, it is called singular.

A key condition for a square matrix A to be invertible is that its determinant must be non-zero, i.e., \( \det(A) \neq 0 \). If \( \det(A) = 0 \), the matrix is singular and its inverse does not exist.

The inverse of a matrix, when it exists, is unique. Matrix inversion has many applications in solving systems of linear equations, linear transformations, and various areas of mathematics, physics, and engineering.

Understanding these properties is fundamental when working with matrix operations, especially in linear algebra and its applications.

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Important Questions from Adjoint and Inverse of a Square Matrix

  1. What should be the value of x so that the matrix \(\left( {\begin{array}{*{20}{c}} 2&4\\ { - 8}&{\rm{x}} \end{array}} \right)\) does not have an inverse?

  2. What is the adjoint of the matrix \(\left( {\begin{array}{*{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) ?

  3. The inverse of the matrix A = \(\left( {\begin{array}{} 1&1&3\\ 1&3&{ - 3}\\ { - 2}&{ - 4}&{ - 4} \end{array}} \right)\) is:

  4. If A is an identity matrix of order 3, then its inverse (A -1 )

  5. What is the inverse of the matrix?

    \(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)

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