Problem 31
Question
If \(A\) is a square matrix, \(B\) is a singular matrix of same order, then for a positive integer \(n,\left(A^{-1} B A\right)^{n}\) equals (A) \(A^{-n} B^{n} A^{n}\) (B) \(A^{n} B^{n} A^{n}\) (C) \(A^{-1} B^{n} A\) (D) \(n\left(A^{-1} B A\right)\)
Step-by-Step Solution
Verified Answer
(A^{-1} B A)^{n} = A^{-1} B^{n} A, option (C).
1Step 1: Understanding the Problem
We are given two matrices, \(A\) and \(B\), where \(A\) is a square matrix, and \(B\) is a singular matrix. We need to find the expression for \((A^{-1} B A)^{n}\) for a positive integer \(n\).
2Step 2: Matrix Power Property
If \(A\) is a matrix, and \(n\) is a positive integer, \((A)^n = A \cdot A \cdot ... \cdot A\) (n times). Thus, we apply this property to the matrix \(A^{-1} B A\).
3Step 3: Expressing \((A^{-1} B A)^n\)
Using the power property, \((A^{-1} B A)^n = (A^{-1} B A) \cdot (A^{-1} B A) \cdot ... \cdot (A^{-1} B A)\), where there are \(n\) factors of \(A^{-1} B A\).
4Step 4: Simplifying \((A^{-1} B A)^n\)
Each multiplication of matrices \(A^{-1} B A\) results in an output identical in form to the input. Therefore, using the associative property of matrices, we can rearrange and simplify:\((A^{-1} B A) (A^{-1} B A) = A^{-1} (BA A^{-1}) B A = A^{-1} B^2 A\).Continuing this pattern:\((A^{-1} B A)^n = A^{-1} B^n A\).
5Step 5: Conclusion
The expression \((A^{-1} B A)^{n}\) simplifies to option (C): \(A^{-1} B^{n} A\).
Key Concepts
Square MatrixSingular MatrixMatrix MultiplicationMatrix Inverse
Square Matrix
A square matrix is one of the fundamental types in matrix algebra, characterized by having the same number of rows and columns. For instance, a 3x3 matrix has three rows and three columns. Square matrices are essential as they allow us to perform operations like calculating determinants, inverses, and powers. The structure is as follows:
- Each row and column are equal in size.
- The elements are usually denoted as \(a_{ij}\), where \(i\) and \(j\) indicate the position within the matrix.
Singular Matrix
A singular matrix is a special type of square matrix that cannot be inverted. This means there is no matrix that you can multiply with it to get an identity matrix. For a matrix \(B\) to be singular, its determinant, a special value calculated from its elements, must be zero:
- Non-zero determinant: matrix is invertible.
- Zero determinant: matrix is singular.
Matrix Multiplication
Matrix multiplication is a key operation in matrix algebra, combining two matrices to form a new matrix. It's not as straightforward as multiplying individual numbers; specific rules must be followed. For matrices \(A\) and \(B\) to be multiplied, the number of columns in \(A\) must equal the number of rows in \(B\). The resulting matrix has dimensions that match the rows of \(A\) and columns of \(B\). Key points include:
- Element \((i, j)\) is calculated by taking the dot product of the \(i\)-th row of \(A\) and the \(j\)-th column of \(B\).
- Matrix multiplication is associative, meaning the grouping of matrices does not affect the result: \((AB)C = A(BC)\).
- Matrix multiplication is generally not commutative: \(AB eq BA\).
Matrix Inverse
The inverse of a matrix is somewhat similar to the inverse of a number; it "undoes" the effect of the matrix. If \(A\) is an invertible matrix, its inverse \(A^{-1}\) satisfies the equation \(AA^{-1} = A^{-1}A = I\), where \(I\) is the identity matrix. The matrix inverse is crucial in solving systems of linear equations, as it can be used to find the solution vector in equations like \(Ax = b\) by manipulating it to \(x = A^{-1}b\). Important aspects include:
- A matrix must be square and non-singular to have an inverse.
- Computing the inverse involves finding a matrix such that when multiplied by the original, results in an identity matrix.
- Not all matrices have inverses; those that can't be inverted are called singular matrices.
Other exercises in this chapter
Problem 29
For each real number \(x\) such that \(-1
View solution Problem 30
The inverse of a skew-symmetric matrix of odd order is (A) a symmetric matrix (B) a skew-symmetric matrix (C) diagonal matrix (D) does not exist
View solution Problem 32
If \(A\) is an invertible matrix, then (A) \(\operatorname{adj} A^{\prime}=(\operatorname{adj} A)^{\prime}\) (B) \(\operatorname{adj} A^{\prime}=\operatorname{a
View solution Problem 33
If \(A\) is a non-singular square matrix of order \(n\), then adj \((\operatorname{adj} A)\) is equal to (A) \(|A|^{n} A\) (B) \(|A|^{n-1} A\) (C) \(|A|^{n-2} A
View solution