\( 1 \leftarrow \quad \) Let \( A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \) and let \( k \) be a scalar. Find a formula that relates \( \operatorname{det}(\mathrm{kA}) \) to \( k \) and \( \operatorname{det}(\mathrm{A}) \). Find \( \operatorname{det}(\mathrm{A}) \). \( \operatorname{det}(\mathrm{A})=\square \) (Simplify your answer.)

To find the relationship between \( \operatorname{det}(kA) \) and \( k \operatorname{det}(A) \), we start by applying the properties of determinants. If \( A \) is a \( 2 \times 2 \) matrix, then \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] The determinant of matrix \( A \) is calculated as follows: \[ \operatorname{det}(A) = ad - bc \] Now, when we scale matrix \( A \) by a scalar \( k \), we have: \[ kA = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix} \] The determinant of the new matrix \( kA \) is: \[ \operatorname{det}(kA) = (ka)(kd) - (kb)(kc) = k^2(ad - bc) = k^2 \operatorname{det}(A) \] Thus, we find that \[ \operatorname{det}(kA) = k^2 \operatorname{det}(A) \] For the problem, we ultimately have: \[ \operatorname{det}(A) = ad - bc \] Hence, filling in the box, the final answer simplifies to: \[ \operatorname{det}(A) = ad - bc \]