Question

Given \(A^{-1}=\left[\begin{array}{rrr}1 & -1 & 3 \\ 2 & 0 & 5 \\ -1 & 1 & 0\end{array}\right]:\) a. Solve the system of equations \(A \mathbf{x}=\left[\begin{array}{r}1 \\ -1 \\\ 3\end{array}\right]\). b. Find a matrix \(B\) such that $$ A B=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{array}\right] $$ c. Find a matrix \(C\) such that \(C A=\left[\begin{array}{rrr}1 & 2 & -1 \\ 3 & 1 & 1\end{array}\right]\)

Short answer

a) \( \mathbf{x} = \begin{bmatrix} 11 \\ 17 \\ -2 \end{bmatrix} \). b) \( B = \begin{bmatrix} 4 & -2 & 1 \\ 7 & -2 & 4 \\ -1 & 2 & -3 \end{bmatrix} \). c) \( C = \begin{bmatrix} 6 & -2 & 13 \\ 4 & -2 & 14 \end{bmatrix} \).

Step-by-step solution

Understanding the System of Equations

The given matrix equation is \( A \mathbf{x} = \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} \). To find \( \mathbf{x} \), we will use the property \( A^{-1} \cdot A = I \), where \( I \) is the identity matrix. Hence, \( \mathbf{x} = A^{-1} \cdot \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} \).

Calculating \( \mathbf{x} \)

To calculate \( \mathbf{x} \), multiply the inverse matrix \( A^{-1} = \begin{bmatrix} 1 & -1 & 3 \ 2 & 0 & 5 \ -1 & 1 & 0 \end{bmatrix} \) by the column vector \( \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} \). Perform the matrix-vector multiplication:\[\mathbf{x} = \begin{bmatrix} 1 \times 1 + (-1) \times (-1) + 3 \times 3 \ 2 \times 1 + 0 \times (-1) + 5 \times 3 \ (-1) \times 1 + 1 \times (-1) + 0 \times 3 \end{bmatrix} = \begin{bmatrix} 1 + 1 + 9 \ 2 + 0 + 15 \ -1 - 1 + 0 \end{bmatrix} = \begin{bmatrix} 11 \ 17 \ -2 \end{bmatrix} \].

Finding Matrix B

The task is to find a matrix \( B \) such that \( A B = D \), where \( D = \begin{bmatrix} 1 & -1 & 2 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} \). To solve for \( B \), multiply both sides by \( A^{-1} \) on the left: \( B = A^{-1} D \).Perform this multiplication: \[B = \begin{bmatrix} 1 & -1 & 3 \ 2 & 0 & 5 \ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + (-1) \times 0 + 3 \times 1 & 1 \times (-1) + (-1) \times 1 + 3 \times 0 & 1 \times 2 + (-1) \times 1 + 3 \times 0 \ 2 \times 1 + 0 \times 0 + 5 \times 1 & 2 \times (-1) + 0 \times 1 + 5 \times 0 & 2 \times 2 + 0 \times 1 + 5 \times 0 \ (-1) \times 1 + 1 \times 0 + 0 \times 1 & (-1) \times (-1) + 1 \times 1 + 0 \times 0 & (-1) \times 2 + 1 \times 1 + 0 \times 0 \end{bmatrix}\]After performing the calculations:\[ B = \begin{bmatrix} 4 & -2 & 1 \ 7 & -2 & 4 \ -1 & 2 & -3 \end{bmatrix} \].

Finding Matrix C

To find matrix \( C \) such that \( C A = \begin{bmatrix} 1 & 2 & -1 \ 3 & 1 & 1 \end{bmatrix} \), multiply both sides on the right by \( A^{-1} \) to isolate \( C \): \[ C = \begin{bmatrix} 1 & 2 & -1 \ 3 & 1 & 1 \end{bmatrix} A^{-1} \].Perform the matrix multiplication:\[C = \begin{bmatrix} 1 & 2 & -1 \ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 3 \ 2 & 0 & 5 \ -1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 2 \times 2 + (-1) \times (-1) & 1 \times (-1) + 2 \times 0 + (-1) \times 1 & 1 \times 3 + 2 \times 5 + (-1) \times 0 \ 3 \times 1 + 1 \times 2 + 1 \times (-1) & 3 \times (-1) + 1 \times 0 + 1 \times 1 & 3 \times 3 + 1 \times 5 + 1 \times 0 \end{bmatrix}\]After completing the calculations:\[ C = \begin{bmatrix} 6 & -2 & 13 \ 4 & -2 & 14 \end{bmatrix} \].

Key concepts

Inverse Matrix

An inverse matrix is a pivotal concept in linear algebra that assists in solving systems of equations and other mathematical problems. When we have a square matrix, denoted as \( A \), its inverse, \( A^{-1} \), satisfies the relation \( A \times A^{-1} = I \), where \( I \) is the identity matrix. The identity matrix is the equivalent of "1" in matrix terminology—it doesn't change any matrix it multiplies when its identity order matches the matrix's order.
The existence of an inverse matrix depends on the matrix being 'invertible' or 'non-singular'. This means the matrix must be square (same number of rows and columns) and have a non-zero determinant. When these conditions are met, multiplying a matrix by its inverse results in the identity matrix.
In the context of solving equations, such as finding \( \mathbf{x} \) in \( A\mathbf{x} = \mathbf{b} \), the inverse is used to isolate \( \mathbf{x} \) by multiplying both sides of the equation by \( A^{-1} \): \( \mathbf{x} = A^{-1}\mathbf{b} \). This technique is very effective for systematically solving systems of linear equations.

Matrix-Vector Multiplication

Matrix-vector multiplication is a basic but crucial operation in linear algebra used for a multitude of applications including solving equations, transformations, and more. Here’s how it works:

• We take a matrix \( A \) with dimensions \( m \times n \) and a vector \( \mathbf{v} \) with dimensions \( n \times 1 \).
• For the operation to be valid, the number of columns in the matrix must match the number of rows in the vector.
• The result of multiplying the matrix \( A \) by the vector \( \mathbf{v} \) is another vector with dimensions \( m \times 1 \).

The multiplication involves taking each row of the matrix \( A \) and performing a dot product with the vector \( \mathbf{v} \). This means multiplying corresponding elements and summing them up to produce a single component of the resultant vector.
In practice, this concept is used to transform vectors and solve matrix equations. For example, when determining the vector \( \mathbf{x} \) in the equation \( A\mathbf{x}=\mathbf{b} \), matrix-vector multiplication allows us to quickly compute the product of an inverse matrix with vector \( \mathbf{b} \), providing the solution vector \( \mathbf{x} \). This process underscores the power of matrix operations in efficiently solving linear equations.

Systems of Linear Equations

Systems of linear equations are essential for modeling numerous real-world situations involving multiple variables. These systems can be written in a compact matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the vector of variables to solve for, and \( \mathbf{b} \) represents the constants.
To solve such a system, especially when dealing with larger matrices, mathematical tools like matrices and their inverses become very useful. One efficient method is to use the inverse matrix, provided \( A \) is invertible, to isolate \( \mathbf{x} \), resulting in \( \mathbf{x} = A^{-1}\mathbf{b} \). This turns a potentially complex system into a straightforward computation.
When the inverse isn't applicable, because the matrix \( A \) is singular or doesn’t exist, other methods such as Gaussian Elimination or utilizing matrix decompositions like LU decomposition might be more appropriate.
Understanding these systems and their solutions aids in various fields like engineering, computer science, and economics, where predicting outcomes based on a set of conditions is crucial.