If $$ A = \begin{pmatrix} 2 & 3 \\ 1 & k \end{pmatrix} $$ and $\det(A) = 7$, find the value of $ k $.

Question

Let $ A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} $. The determinant of $ A^3 $ is:

Answer 1

The determinant of matrix $A$ is calculated using the formula: $\det(A) = ad - bc$.

For the matrix $ A = \begin{pmatrix} 2 & 3 \\ 1 & k \end{pmatrix} $, the elements are $ a = 2 $, $ b = 3 $, $ c = 1 $, and $ d = k $.

Substituting these values into the determinant formula yields: $\det(A) = (2)(k) - (3)(1) = 2k - 3$.

Given that $\det(A) = 7$, we set up the equation: $ 2k - 3 = 7 $.

To solve for $ k $, we first add 3 to both sides of the equation: $ 2k = 10 $.

Finally, we divide both sides by 2: $ k = 5 $.

The resulting value for $ k $ is 5.

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