To check if a matrix $A$ is invertible, we need to ensure that the determinant is non-zero. Alternatively, we can perform Gaussian elimination to bring the matrix to its Reduced Row Echelon Form (RREF). If the matrix can be reduced to the identity matrix, then $A$ is invertible. Otherwise, it is not.

The matrix $A$ given is a 3x3 matrix:

1 & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 4 & 6 \end{bmatrix}$$ Let's augment $$A$$ with the identity matrix $$I$$, and then attempt to reduce $$[A|I]$$ to row-echelon form. $$[A|I] = \begin{bmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 2 & 4 & 6 & 0 & 0 & 1 \end{bmatrix}$$ I'll perform the row-reduction steps next.The augmented matrix $$[A|I]$$ reduces to: $$\begin{bmatrix} 1 & 0 & -1 & 0 & \frac{2}{3} & -\frac{5}{6} \\ 0 & 1 & 2 & 0 & -\frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 0 & 1 & 0 & -\frac{1}{2} \end{bmatrix}$$ Notice that the third row contains all zeros in the first three columns, indicating that the original matrix $$A$$ cannot be reduced to the identity matrix on the left side. Therefore, $$A$$ is **not invertible**. ### Summary: - The matrix $$A$$ is not invertible because the row-reduction process fails to produce the identity matrix on the left side. - This result also aligns with the fact that the determinant of $$A$$ would be zero. Would you like more details or have any questions? Here are 5 related questions for further exploration: 1. How do you compute the determinant of a matrix, and why does a zero determinant imply non-invertibility? 2. What is the geometric interpretation of a non-invertible matrix? 3. Can a matrix with linearly dependent rows or columns be invertible? 4. How does Gaussian elimination help in finding the inverse of a matrix? 5. What is the significance of the pivot columns in the RREF process? **Tip:** If a matrix has two identical or proportional rows, it is always non-invertible.