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

When you multiply a matrix \( A \) by a scalar \( k \), each element of \( A \) is multiplied by \( k \). This means that if \( A \) is a 2x2 matrix, the determinant of \( kA \) can be found using the formula \( \operatorname{det}(kA) = k^n \cdot \operatorname{det}(A) \), where \( n \) is the size of the matrix (in this case, \( 2 \)). So, \( \operatorname{det}(kA) = k^2 \cdot \operatorname{det}(A) \). For the determinant of \( A \) given as \( A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \), the determinant is calculated using the formula \( \operatorname{det}(A) = ad - bc \). Therefore, if you want to express \( \operatorname{det}(kA) \), it becomes \( \operatorname{det}(kA) = k^2(ad - bc) \). Voilà! You’ve simplified the relationship between the determinant of a scaled matrix and the original matrix.