Problem 38
Question
\(A\) and \(B\) are two non-singular matrices of the same order such that \(A^{n}=I\) for some positive integer \(n>1\). Then, \(B A^{n-1} B^{-1}-B A^{-1} B^{-1}\) (A) is a null matrix (B) is an identity matrix (C) a singular matrix (D) None of these
Step-by-Step Solution
Verified Answer
The expression simplifies to the null matrix.
1Step 1: Understanding the Given Information
We are given that matrix \(A\) is non-singular, meaning it has an inverse, and \(A^n = I\), where \(I\) is the identity matrix and \(n > 1\). This implies that when \(A\) is raised to the power of \(n\), it results in the identity matrix.
2Step 2: Simplifying the Expression
We need to simplify the expression \(B A^{n-1} B^{-1} - B A^{-1} B^{-1}\). Since \(A^n = I\), we have \(A^n = A \cdot A^{n-1} = I\), implying that \(A^{n-1} = A^{-1}\).
3Step 3: Substituting Into the Expression
Substitute \(A^{n-1} = A^{-1}\) into the expression \(B A^{n-1} B^{-1} - B A^{-1} B^{-1}\). This gives us \(B A^{-1} B^{-1} - B A^{-1} B^{-1}\).
4Step 4: Conclusion After Simplifying
Now, the expression becomes \(0\) because the two terms \(B A^{-1} B^{-1}\) are equal and they subtract to give the zero matrix, also known as the null matrix.
Key Concepts
Non-Singular MatricesIdentity MatrixInverse Matrix
Non-Singular Matrices
In the world of matrix algebra, a non-singular matrix is quite important. Simply put, a non-singular matrix is a square matrix that has an inverse. It means, if you multiply a non-singular matrix by its inverse, you get the identity matrix as the result. So, if we have a matrix denoted by \(A\), and we find another matrix called \(A^{-1}\), such that \(A \times A^{-1} = I\), then \(A\) is non-singular.
This concept is essential because non-singular matrices are used to solve linear equations; an operation not possible with singular matrices since they don't have inverses. Furthermore, non-singular matrices also ensure the determinant is not zero. This characteristic is what makes them invertible, as a zero determinant implies singularity, and thus a lack of an inverse.
Moreover, non-singular matrices are crucial in understanding systems of linear equations, transformations, and in every aspect where matrix multiplicity and matrix equations are involved.
This concept is essential because non-singular matrices are used to solve linear equations; an operation not possible with singular matrices since they don't have inverses. Furthermore, non-singular matrices also ensure the determinant is not zero. This characteristic is what makes them invertible, as a zero determinant implies singularity, and thus a lack of an inverse.
Moreover, non-singular matrices are crucial in understanding systems of linear equations, transformations, and in every aspect where matrix multiplicity and matrix equations are involved.
Identity Matrix
The identity matrix is a unique backdrop in matrix algebra. It's like the number '1' in multiplication. When a matrix is multiplied by an identity matrix, it remains unchanged. An identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. For example, a 3x3 identity matrix looks like this:
\[\begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]
When any square matrix \(A\) is multiplied by an identity matrix of the same order, the original matrix \(A\) is the product: \( A \times I = A \) and \( I \times A = A \).
As seen in the original exercise, when matrix \(A\) is raised to a power \(n > 1\) to become an identity matrix \(I\), it signifies a very special condition: it signifies that a multiple application of the transformation represented by \(A\) eventually returns the system to its starting condition, much like revolving through a complete cycle in periodic functions.
\[\begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]
When any square matrix \(A\) is multiplied by an identity matrix of the same order, the original matrix \(A\) is the product: \( A \times I = A \) and \( I \times A = A \).
As seen in the original exercise, when matrix \(A\) is raised to a power \(n > 1\) to become an identity matrix \(I\), it signifies a very special condition: it signifies that a multiple application of the transformation represented by \(A\) eventually returns the system to its starting condition, much like revolving through a complete cycle in periodic functions.
Inverse Matrix
An inverse matrix is essentially what "undoes" the multiplication done by the original matrix. If a matrix \(A\) is multiplied by an inverse \(A^{-1}\), the result is an identity matrix \(I\). This is a crucial aspect when solving equations involving matrices. The ability to "invert" or undo operations is what allows systems of equations to be solved explicitly where matrix coefficients are involved.
To find an inverse of a matrix, the matrix must be non-singular (have a non-zero determinant). The inverse is denoted by a symbol \(A^{-1}\), and their relationship is expressed as \(A \times A^{-1} = I\) and \(A^{-1} \times A = I\).
In the original exercise, the property \(A^n = I\) was crucial. It inferred that \(A^{n-1} = A^{-1}\), because multiplying both sides by \(A\) grants the identity matrix, thus proving \(A^{-1}\) in practical use where \(A\) repeated \(n\) times equals the neutral element, the identity matrix, much like an eventual equilibrium.
To find an inverse of a matrix, the matrix must be non-singular (have a non-zero determinant). The inverse is denoted by a symbol \(A^{-1}\), and their relationship is expressed as \(A \times A^{-1} = I\) and \(A^{-1} \times A = I\).
In the original exercise, the property \(A^n = I\) was crucial. It inferred that \(A^{n-1} = A^{-1}\), because multiplying both sides by \(A\) grants the identity matrix, thus proving \(A^{-1}\) in practical use where \(A\) repeated \(n\) times equals the neutral element, the identity matrix, much like an eventual equilibrium.
Other exercises in this chapter
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) \(\
View solution Problem 37
If \(A^{3}=0\) and \(A^{n} \neq I\) for \(n=1,2\) then \((I-A)^{-1}\) is (A) \(I+A\) (B) \(I+A+A^{2}\) (C) \(I-A+A^{2}\) (D) None of these
View solution Problem 40
A skew-symmetric matrix \(A\) satisfies the relation \(A^{2}+\) \(I=0\), where \(I\) is a unit matrix. Then, \(A\) is (A) Idempotent matrix (B) Orthogonal matri
View solution Problem 41
Let \(A\) be an \(n \times n\) matrix such that \(A^{n}=\alpha A\), where \(\alpha\) is a real number different from 1 and \(-1\). Then, the matrix \(A+I_{n}\)
View solution