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The first part of the statement deals with the inverse of matrix A. The inverse of a matrix A exists if it is a square matrix (i.e., it has the same number of rows and columns) and its determinant is non-zero. A matrix with no inverse is called a singular matrix. In this exercise, we are given that \(A^{-1}\) does not exist.

The existence of an inverse has a significant impact on the solution of a system of linear equations, AX = B. If there exists a unique solution of AX = B, then it can be found by left-multiplying both sides of the equation with the inverse of A such that \(A^{-1} (A X) = A^{-1} B\) or simply X = \(A^{-1} B\). Since \(A^{-1}\) does not exist, we cannot use this method to find the unique solution of AX = B. The system AX = B can have either no solution or infinitely many solutions depending on the setup of the linear system.

According to the given statement, we need to show that if \(A^{-1}\) does not exist, then the system AX = B does not have a unique solution. This statement is true. Explanation: Let's consider two possibilities for a system of linear equations without a unique solution - no solutions or infinitely many solutions. 1. No solution: This occurs when the system of linear equations is inconsistent, which means that the rows of the augmented matrix ([A|B]) represent contradictory equations. In this case, the system cannot have a unique solution. 2. Infinitely many solutions: This occurs when the system of linear equations has dependent equations, and one or more variables can take any values. It means there is a free variable in the echelon form of the coefficient matrix. In this case, the system cannot have a unique solution. Since we have shown that AX = B either has no solution or infinitely many solutions if \(A^{-1}\) does not exist, the statement is considered true.