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Chapter 5: Problem 36

If \(A\) is a non-singular matrix such that \(A A^{\prime}=A^{\prime} A\) and \(B=A^{-1} A^{\prime}\), then \(B B^{\prime}\) is (A) \(I\) (B) \(B^{-1}\) (C) \(\left(B^{-1}\right)^{\prime}\) (D) None of these

Short Answer

Expert verified
The answer is (A) \(I\).

Step by step solution

01

Understand Matrices and Transposes

The problem gives us a non-singular matrix \(A\) and uses the fact that \(A A^{\prime} = A^{\prime} A\). A non-singular matrix is invertible, and \(A^{\prime}\) denotes the transpose of \(A\). These properties help establish that the matrices commute.
02

Define Matrix B

Matrix \(B\) is defined as \(B = A^{-1} A^{\prime}\). This expression depends on the inverse and transpose of \(A\). The matrix \(B\) essentially transforms \(A\) while maintaining its core properties.
03

Express BB'

We need to find \(B B^{\prime}\). First, calculate the transpose of \(B\), which is \(B^{\prime} = (A^{-1} A^{\prime})^{\prime}\). This simplifies to \((A^{\prime})^{\prime} (A^{-1})^{\prime} = A A^{-1}\), as the transpose of a product of matrices reverses the order and transposes individually each matrix.
04

Simplify B'B

Now, we calculate \(B B^{\prime}\) by substituting back: \(B B^{\prime} = (A^{-1} A^{\prime})(A A^{-1}) = A^{-1} (A^{\prime} A) A^{-1}\). Since \(A A^{\prime} = A^{\prime} A\), this further simplifies to \(A^{-1} A A^{-1} = I\), where \(I\) is the identity matrix.
05

Conclusion

Based on the simplifications, we conclude that \(B B^{\prime} = I\). Hence, the correct option is (A) \(I\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Inverse
The concept of a matrix inverse is crucial in understanding linear algebra. For a given square matrix \(A\), its inverse \(A^{-1}\) is the matrix that, when multiplied by \(A\), yields the identity matrix \(I\). This is symbolized as \(AA^{-1} = I\). The identity matrix \(I\) acts like the number 1 in matrix multiplication, meaning any matrix multiplied by \(I\) remains unchanged.
To find an inverse, a matrix must be non-singular (i.e., it must have a non-zero determinant). If you have a singular matrix, it does not have an inverse. Calculating the inverse involves complex methods such as Gaussian elimination or using the adjugate and determinant of the matrix. However, understanding the essence of an inverse鈥攅specially in operations like transformations and solving systems of equations鈥攊s key. These operations underline the importance of matrix inverses in mathematical applications and their relevance in achieving matrix commutativity as seen in the given problem.
Non-Singular Matrix
A non-singular matrix, also known as an invertible matrix, is a matrix that possesses an inverse. The defining property of a non-singular matrix is that its determinant is not zero. This distinction is vital because only non-singular matrices can be inverted.
  • If a matrix is non-singular, it has full rank, implying that its rows and columns are linearly independent.
  • Because it is invertible, multiplying a non-singular matrix by its inverse results in the identity matrix.
  • It allows for transformations and solving linear systems, preserving the essence of the original data when transformations are applied, as seen through the exercise solution.
Understanding non-singular matrices is fundamental when dealing with complex matrix operations and is crucial for ensuring that given matrices retain their mathematical properties across calculations.
Commutative Property of Matrices
In general, matrix multiplication is not commutative, which means that \(AB eq BA\) for two matrices \(A\) and \(B\). However, in some special cases, matrices can commute. The exercise problem gives us a scenario where \(A A^{\prime} = A^{\prime} A\), indicating that the transpose of the matrix commutes with the matrix itself.
  • The commutative property simplifies calculations as it allows rearranging the order of multiplication without changing the result.
  • In this problem, this property enables simplification when computing expressions like \(B B^{\prime}\).
Commutativity in the realm of matrices is rare, but understanding the exceptional cases helps in efficiently solving matrix equations and contributes to understanding deeper matrix algebraic structures.

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Most popular questions from this chapter

Let \(\mathrm{A}\) and \(\mathrm{B}\) be two symmetric matrices of order \(3 .\) \([2011]\) Statement 1: \(A(B A)\) and \((A B) A\) are symmetric matrices. Statement 2: \(A B\) is symmetric matrix if matrix multiplication of \(A\) and \(B\) is commutative. (A) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (B) Statement 1 is true, Statement 2 is false. (C) Statement 1 is false, Statement 2 is true. (D) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

If \(A\) and \(B\) are symmetric matrices and \(A B=B A\), then \(A^{-1} B\) is a (A) symmetric matrix (B) skew-symmetric matrix (C) identity matrix (D) None of these

If the product of the matrix \(B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]\) with a matrix \(A\) has inverse \(C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\\ 2 & 0 & 2\end{array}\right]\), then \(A^{-1}\) equals (A) \(\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6\end{array}\right]\) (B) \(\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]\) (C) \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\) (D) \(\left[\begin{array}{ccc}-3 & -3 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)

Which of the following is true? (A) Transpose of an orthogonal matrix is also orthogonal (B) Every orthogonal matrix is non-singular (C) Product of the two orthogonal matrices is also orthogonal (D) Inverse of an orthogonal matrix is also orthogonal

Suppose, \(a, b, c\) are real numbers such that \(a b c=1\). If the matrix \(A=\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]\) is such that \(A^{\prime} A=I\), then the value of \(a^{3}+b^{3}+c^{3}\) is (A) 1 (B) 2 (C) 3 (D) 4
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