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If \(A, B,\) and \(C\) are \(n \times n\) invertible matrices, does the equation \(C^{-1}(A+X) B^{-1}=I_{n}\) have a solution, \(X ?\) If so, find it.

Short Answer

Expert verified
Yes, the solution is \(X = CB - A\).

Step by step solution

01

Understanding the Equation

We are given the equation \(C^{-1}(A+X)B^{-1}=I_{n}\), where \(I_{n}\) is the \(n \times n\) identity matrix. The goal is to find if there is a matrix \(X\) that satisfies this equation.
02

Simplifying the Equation

To isolate \(X\), we need to manipulate the equation. Start by multiplying both sides of the equation by \(C\) and \(B\) to get rid of the inverses. Multiplying by \(C\) on the left and \(B\) on the right yields: \((A+X)= C I_{n} B = CB\). Thus, \(A + X = CB\).
03

Solving for X

From the equation \(A + X = CB\), solve for \(X\) by subtracting \(A\) from both sides, yielding \(X = CB - A\). This gives us the solution for \(X\).
04

Verify the Solution

Plug \(X = CB - A\) back into the original equation. Replace \(X\) in the equation with \(CB - A\): \(C^{-1}(A + (CB - A))B^{-1} = C^{-1}(CB)B^{-1} = I_{n}\). Since \(C^{-1}C = I_{n}\) and \(BB^{-1} = I_{n}\), the right side simplifies to \(I_{n}\), verifying that this is indeed the solution.

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Key Concepts

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

Invertible Matrices
Invertible matrices, also known as nonsingular or nondegenerate matrices, hold special properties that make them quite useful in algebra. A square matrix is invertible if there exists another matrix, called its inverse, that can be multiplied with the original to produce the identity matrix. In more formal terms, a matrix \( A \) is invertible if there exists a matrix \( A^{-1} \) such that:
  • \( A \times A^{-1} = I_{n} \)
  • \( A^{-1} \times A = I_{n} \)
Here, \( I_{n} \) signifies an identity matrix with the same dimensions as \( A \). This property ensures that operations involving inverse matrices can simplify equations significantly, as they allow for methods like cancellation, useful in finding solutions to matrix equations. Not every matrix is invertible; only those with a non-zero determinant can be inverted.
Identity Matrix
The identity matrix is a pivotal concept in linear algebra and matrix operations. Denoted typically as \( I_{n} \) for an \( n \times n \) matrix, the identity matrix has 1s on its main diagonal and 0s elsewhere:
  • \( I_{n} = \begin{pmatrix} 1 & 0 & \cdots & 0 \ 0 & 1 & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & 1 \end{pmatrix} \)
The identity matrix acts as the multiplicative identity for matrix multiplication, much like the number 1 works in arithmetic. This means for any matrix \( A \) (of compatible dimensions), \( A \times I_{n} = A \) and \( I_{n} \times A = A \). In the context of matrix equations, the identity matrix is crucial, providing a baseline for solutions, especially when dealing with invertible matrices, as seen in the original equation \( C^{-1}(A+X)B^{-1}=I_{n} \). It helps verify solutions, ensuring that after manipulation, the output aligns perfectly with the identity matrix's properties.
Matrix Manipulation
Matrix manipulation encompasses various operations and transformations one can apply to matrices to solve equations or simplify expressions. Common manipulations include addition, subtraction, multiplication, and scalar operations. In the context of more complex equations, manipulating involves steps like:
  • Distributing matrices over addition or subtraction, similar to algebra.
  • Applying row or column transformations to simplify matrices.
  • Using properties of the identity matrix to cancel out terms.
  • Utilizing the inverse of a matrix to isolate variables or simplify.
The operation often requires an understanding of specific rules governing matrices, such as the non-commutative nature of matrix multiplication—that is, \( A \times B \) may not necessarily equal \( B \times A \). In the exercise at hand, matrix manipulation is used to solve \( C^{-1}(A+X)B^{-1}=I_{n} \), isolating \( X \) through strategically multiplying out and rearranging terms to arrive at \( X = CB - A \).
Matrix Inversion
Matrix inversion is a vital process in linear algebra, allowing for the resolution of matrix equations and systems of linear equations. An inverse matrix effectively "undoes" the multiplication with its original matrix, akin to the inverse operations seen in basic arithmetic. To find the inverse of a matrix \( A \), denoted \( A^{-1} \), one must ensure:
  • The matrix is square (same number of rows and columns).
  • The determinant of the matrix is non-zero.
Various methods exist to calculate an inverse, such as the Gaussian elimination or the adjugate method. Once computed, this matrix satisfies the conditions \( A \times A^{-1} = I_{n} \) and \( A^{-1} \times A = I_{n} \), where \( I_{n} \) is the identity matrix. Inverting matrices allows complex systems and equations to be simplified and solved by making critical elements cancel out. In the step-by-step solution provided, matrix inversion is used to isolate \( X \) by eliminating other components of the equation through calculated inverse multiplication.

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Most popular questions from this chapter

Determine which sets in Exercises \(15-20\) are bases for \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\) . Justify each answer. $$ \left[\begin{array}{r}{-4} \\ {6}\end{array}\right],\left[\begin{array}{r}{2} \\\ {-3}\end{array}\right] $$

In Exercises 1 and \(2,\) find the vector \(\mathbf{x}\) determined by the given coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) and the given basis \(\mathcal{B}\) . Illustrate your answer with a figure, as in the solution of Practice Problem \(2 .\) $$\mathcal{B}=\left\\{\left[\begin{array}{l}{1} \\\ {1}\end{array}\right],\left[\begin{array}{r}{2} \\\ {-1}\end{array}\right]\right\\},[\mathbf{x}]_{\mathcal{B}}=\left[\begin{array}{l}{3} \\\ {2}\end{array}\right]$$

In Exercises 17 and \(18,\) mark each statement True or False. Justify each answer. Here \(A\) is an \(m \times n\) matrix. a. If \(\mathcal{B}=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is a basis for a subspace \(H\) and if \(\mathbf{x}=c_{1} \mathbf{v}_{1}+\cdots+c_{p} \mathbf{v}_{p},\) then \(c_{1}, \ldots, c_{p}\) are the coordinates of \(\mathbf{x}\) relative to the basis \(\mathcal{B}\) . b. Each line in \(\mathbb{R}^{n}\) is a one-dimensional subspace of \(\mathbb{R}^{n}\) . c. The dimension of Col \(A\) is the number of pivot columns of \(A .\) d. The dimensions of Col \(A\) and Nul \(A\) add up to the number of columns of \(A\) . e. If a set of \(p\) vectors spans a \(p\) -dimensional subspace \(H\) of \(\mathbb{R}^{n},\) then these vectors form a basis for \(H .\)

In Exercises 31–36, respond as comprehensively as possible, and justify your answer. Suppose \(F\) is a \(5 \times 5\) matrix whose column space is not equal to \(\mathbb{R}^{5} .\) What can you say about Nul \(F ?\)

In Exercises \(3-6,\) the vector \(\mathbf{x}\) is in a subspace \(H\) with a basis \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\} .\) Find the \(\mathcal{B}\) -coordinate vector of \(\mathbf{x} .\) $$ \mathbf{b}_{1}=\left[\begin{array}{r}{1} \\ {5} \\ {-3}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{-3} \\ {-7} \\ {5}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{4} \\ {10} \\ {-7}\end{array}\right] $$

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