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Magazine Chapter 4: Problem 54
\(A\) and \(B\) are \(n \times n\) matrices. If \(B\) is invertible, prove that \(\operatorname{det}\left(B^{-1} A B\right)=\operatorname{det}(A)\)
Short Answer
Expert verified
\(\operatorname{det}(B^{-1}AB) = \operatorname{det}(A)\) by determinant properties.
Step by step solution
01
Understanding Determinant Properties
The determinant of a matrix and its relation to products and inverses is crucial here. We need to recall that for any invertible matrix \(B\) of the same size, \(\operatorname{det}(B^{-1}) = \frac{1}{\operatorname{det}(B)}\) and for any two matrices \(X\) and \(Y\) of the same size, \(\operatorname{det}(XY) = \operatorname{det}(X) \times \operatorname{det}(Y)\).
02
Computing the Determinant of a Product
Given the expression \(\operatorname{det}(B^{-1}AB)\), we recognize that it is a product of three matrices: \(B^{-1}\), \(A\), and \(B\). So, we can express it using the properties of determinants: \(\operatorname{det}(B^{-1}AB) = \operatorname{det}(B^{-1}) \times \operatorname{det}(A) \times \operatorname{det}(B)\).
03
Substitute Inverse's Determinant
Next, substitute the expression for \(\operatorname{det}(B^{-1})\) from the determinant properties: \(\operatorname{det}(B^{-1}AB) = \frac{1}{\operatorname{det}(B)} \times \operatorname{det}(A) \times \operatorname{det}(B)\).
04
Simplifying the Expression
The \(\operatorname{det}(B)\) in the numerator and denominator cancel each other: \(\frac{1}{\operatorname{det}(B)} \times \operatorname{det}(B) = 1\). Thus, we have \(\operatorname{det}(B^{-1}AB) = \operatorname{det}(A)\).
05
Conclusion
Through these steps, we have shown that \(\operatorname{det}(B^{-1}AB) = \operatorname{det}(A)\) when \(B\) is invertible, thereby proving the statement.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Invertible Matrices
An invertible matrix, often referred to as a non-singular or non-degenerate matrix, is a square matrix that has an inverse. That is, for a matrix \( B \), there exists another matrix \( B^{-1} \) such that when you multiply them together, you get the identity matrix.
- If \( B \) is \( n \times n \), then \( B \times B^{-1} = B^{-1} \times B = I_n \), where \( I_n\) is the \( n \times n \) identity matrix.
- An important condition for a matrix to be invertible is that its determinant should not be zero. If \( \operatorname{det}(B) eq 0 \), then \( B \) is invertible.
Determinant Properties
The determinant is a special number that can be computed from a square matrix. It provides meaningful insights into the matrix's properties, such as whether the matrix is invertible. Let's discuss the two key determinant properties relevant to our exercise:
- For any invertible matrix \( B \), the determinant of its inverse is given by \( \operatorname{det}(B^{-1}) = \frac{1}{\operatorname{det}(B)} \).
- The determinant of a product of two matrices, \( X \) and \( Y \), satisfies the property \( \operatorname{det}(XY) = \operatorname{det}(X) \times \operatorname{det}(Y) \).
Matrix Multiplication
Matrix multiplication is a process of multiplying two matrices together to form another matrix. For our discussion:
- Only matrices that have compatible dimensions can be multiplied. For example, you can multiply \( m \times n \) matrix \( A \) with \( n \times p \) matrix \( B \), resulting in an \( m \times p \) matrix.
- Multiplication of matrices is associative, implying \((B^{-1}A)B = B^{-1}(AB)\), but it's not commutative, meaning \( AB eq BA \).
Matrix Inverses
The inverse of a matrix reverses the effect of the original matrix. Consider a situation when you have an invertible matrix \( B \). Its inverse \( B^{-1} \) has crucial applications:
- If you multiply \( B \) by \( B^{-1} \), the product is the identity matrix \( I \).
- The row operations leading to forming \( B^{-1} \) include swapping rows, multiplying a row by a scalar, and adding a multiple of one row to another. Each of these does not alter the determinant's property of a matrix being invertible.
