Question

What is the solution of the system of equations \(A X=0\) if \(A^{-1}\) exists? Discuss the solution of \(A X=0\) if \(A^{-1}\) does not exist.

Short answer

If \(A^{-1}\) exists, the only solution is \(X = 0\). If \(A^{-1}\) does not exist, there are either no solutions or infinitely many solutions, but \(X = 0\) is always a solution.

Step-by-step solution

Understand the given system

The problem involves solving the system of linear equations represented by the matrix equation \(A X = 0\), where \(A\) is a matrix and \(X\) is the vector of variables.

Check if \(A^{-1}\) exists

To determine the type of solution, first check if the inverse of matrix \(A\) exists. A matrix \(A\) will have an inverse \(A^{-1}\) if and only if \(A\) is non-singular, which means its determinant is non-zero.

Consider the case when \(A^{-1}\) exists

If \(A^{-1}\) exists, we can multiply both sides of the equation \(A X = 0\) by \(A^{-1}\). This yields: \[A^{-1} A X = A^{-1} 0\]Since \(A^{-1} A\) is the identity matrix \(I\), we get: \[I X = 0\]Which simplifies to: \[X = 0\]So, the only solution in this case is the trivial solution \(X = 0\).

Discuss the case when \(A^{-1}\) does not exist

If \(A^{-1}\) does not exist, then \(A\) is singular. This means the determinant of \(A\) is zero, and \(A\) does not have full rank. In this case, the system \(A X = 0\) can have either no solutions or infinitely many solutions, but it always has the trivial solution \(X = 0\). To determine the exact nature of the solutions, further analysis (such as row reducing \(A\) to its echelon form) would be required.

Key concepts

Matrix Inverse

In linear algebra, the inverse of a matrix, denoted as \(A^{-1}\), is a matrix that, when multiplied with the original matrix \(A\), yields the identity matrix \(I\). Mathematically, it is represented as: \ \ \[A A^{-1} = A^{-1} A = I\] \ \ For a matrix to have an inverse, it must be a square matrix (same number of rows and columns) and non-singular (more on this later). Not all matrices have inverses, and determining if an inverse exists is crucial in solving linear systems. When we have \(A X = 0\) and \(A^{-1}\) exists, we can multiply both sides by \(A^{-1}\) to isolate \(X\). This yields the trivial solution \(X = 0\).

Determinant

The determinant of a matrix is a special number calculated from its elements and helps determine whether the matrix is invertible. For a square matrix \(A\), the determinant is denoted as \(\text{det}(A)\). If \(\text{det}(A) eq 0\), the matrix is non-singular and an inverse exists. Conversely, if \(\text{det}(A) = 0\), the matrix is singular, and an inverse does not exist. \ \ To solve \(A X = 0\), we first check the determinant of \(A\). If it's non-zero, we proceed knowing \(A^{-1}\) exists. In our context, calculating the determinant helps to understand the matrix's properties without needing its elements' explicit values.

Singular Matrix

A singular matrix is a square matrix whose determinant is zero \(\text{det}(A) = 0\). This implies that the matrix does not have an inverse. In the context of the linear system \(A X = 0\), a singular matrix suggests that there might be no unique solution. When \(A\) is singular: \ \

• \(A X = 0\) might have no solutions
• There could be infinitely many solutions
• It always includes the trivial solution \(X = 0\)

\ To determine if a matrix is singular, check its determinant. If zero, the matrix is singular, meaning further steps such as row reduction are needed to understand the solution space better.

Trivial Solution

In the context of solving \(A X = 0\), the trivial solution refers to \(X = 0\). This simple situation occurs when every variable in the vector \(X\) is zero. It is called 'trivial' because it's the most straightforward solution and always satisfies \(A X = 0\), regardless of \(A\). \ \ When \(A^{-1}\) exists, the only solution to \(A X = 0\) is the trivial solution. However, when \(A\) is singular \((\text{det}(A) = 0)\), there can be more than just the trivial solution. In such cases, analyzing the matrix's rank and performing row reduction can help identify other potential solutions.