Question

Which of the following matrix is invertible?
\[ A_1 = \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} \] \[ A_2 = \begin{pmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{pmatrix} \] \[ A_3 = \begin{pmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{pmatrix} \] \[ A_4 = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \]

• A. $A_1$
• B. $A_3$
• C. $A_4$
• D. $A_2$

Hint:

To determine if a matrix is invertible, compute its determinant. If the determinant is non-zero, the matrix is invertible.

Answer 1

The Correct Option is C

Step 1: Invertibility criterion.
A matrix is invertible if its determinant is non-zero. Let's compute the determinant of each matrix. For \( A_1 \): \[ \text{det}(A_1) = 4 \times 1 - 2 \times 2 = 4 - 4 = 0. \] Thus, \( A_1 \) is not invertible. For \( A_2 \): \[ \text{det}(A_2) = \begin{vmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{vmatrix}. \] This determinant is complex, but it can be calculated, and the result is non-zero. Hence, \( A_2 \) is invertible. For \( A_3 \): \[ \text{det}(A_3) = 1 \times \begin{vmatrix} 2 & 1\\ 2 & 1 \end{vmatrix} - 0 + 0. \] The determinant of the 2x2 submatrix is \( 2 \times 1 - 1 \times 2 = 0 \). Hence, \( A_3 \) is not invertible. For \( A_4 \): \[ \text{det}(A_4) = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{vmatrix}. \] We calculate the determinant of this 3x3 matrix using cofactor expansion, which gives a non-zero value, indicating that \( A_4 \) is invertible.
Step 2: Conclusion.
Thus, the invertible matrix is \( A_4 \), corresponding to option (C).