The statement "If $A^{-1}$ exists, then ( \det(A) = 0 " is false. Here's why:

Correct Statement:

If $A^{-1}$ exists, then $\det(A) \neq 0$.

Explanation:

1. Invertibility and Determinant: For a square matrix $A$ to have an inverse $A^{-1}$, it must be invertible or nonsingular. A fundamental property of matrices is that a matrix is invertible if and only if its determinant is non-zero. This is because:

   $A^{-1} \text{ exists } \iff \det(A) \neq 0$

2. Singular Matrices: If $\det(A) = 0$, the matrix $A$ is called singular, and it does not have an inverse. This occurs because the determinant being zero implies that the matrix has linearly dependent rows or columns, making the system of equations associated with the matrix unsolvable in a unique way.

3. Geometrically: In 2D or 3D, the determinant of a matrix can be interpreted as the scaling factor of the transformation described by the matrix. A zero determinant means that the transformation squashes some dimension, collapsing the space into a lower dimension (hence, non-invertible).

Conclusion:

Thus, the correct implication is:

$A^{-1} \text{ exists} \implies \det(A) \neq 0$

Would you like further details or have any questions?

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Related Questions:

1. What is the relationship between the determinant and eigenvalues of a matrix?
2. How do we compute the determinant of a matrix in practice?
3. Can a matrix with a non-zero determinant ever be non-invertible?
4. How is the inverse of a matrix calculated, and how does it involve the determinant?
5. What happens if $\det(A) = 0$ for a system of linear equations?

Tip:

To quickly check if a matrix is invertible, calculate its determinant. If the determinant is zero, the matrix is not invertible.