A matrix is invertible if its determinant is non-zero. The determinant of $$C$$ is $$(k-2)(k+2) - (3)(3) = k^2 - 4 - 9 = k^2 - 13$$. For the matrix to be invertible, we must have $$k^2 - 13 \neq 0$$, which means $$k^2 \neq 13$$. Therefore, $$k \neq \pm \sqrt{13}$$.

The inverse of a $$2 \times 2$$ matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ is $$\frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$. First, calculate the determinant of $$A$$: $$\det(A) = (2)(3) - (5)(1) = 6 - 5 = 1$$. Then, apply the formula for the inverse: $$A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$$

The area of the parallelogram formed by the column vectors of a matrix is the absolute value of its determinant. Since $$\det(A) = 4$$, the area is $$|4| = 4$$. Because the determinant is non-zero, the matrix must have an inverse. There are many matrices with a determinant of 4, not just the one given. The product of a matrix and its inverse is always the identity matrix, not a matrix with the determinant on the diagonal.

The determinant is $$(\ln 2)(\ln 9) - (\ln 3)(\ln 4)$$. Using logarithm properties, we can rewrite this as $$(\ln 2)(\ln 3^2) - (\ln 3)(\ln 2^2)$$. This simplifies to $$(\ln 2)(2 \ln 3) - (\ln 3)(2 \ln 2)$$. Both terms are equal to $$2(\ln 2)(\ln 3)$$, so their difference is 0. Therefore, the determinant is 0.

If the determinant of a $$2 \times 2$$ matrix is zero, the column (and row) vectors are linearly dependent, which means they are parallel (or collinear). In this case, $$\vec{v}_2 = -4\vec{v}_1$$. Perpendicular vectors would have a dot product of zero. The magnitudes are different, and they do not necessarily form a square.

This question tests AP Precalculus skills involving matrices, specifically determining invertibility by calculating determinants of multiple matrices. A matrix is invertible if and only if its determinant is non-zero, requiring separate calculations for each matrix. In this problem, for A = [[1,2],[3,4]], det(A) = 1(4) - 2(3) = 4 - 6 = -2 ≠ 0, so A is invertible; for B = [[2,4],[1,2]], det(B) = 2(2) - 4(1) = 4 - 4 = 0, so B is not invertible. Choice A is correct because only matrix A has a non-zero determinant. Choice B incorrectly identifies B as invertible, choice C claims both are invertible despite B having det = 0, and choice D claims neither is invertible despite A having det ≠ 0. To help students: calculate determinants systematically for each matrix, recognize that proportional rows (in B, row 1 = 2×row 2) always yield det = 0, and practice identifying invertible vs. non-invertible matrices quickly.

This question tests AP Precalculus skills involving matrices, specifically calculating the inverse of a 2x2 matrix using the standard formula. The inverse formula for matrix [[a,b],[c,d]] is (1/(ad-bc))[[d,-b],[-c,a]], requiring careful attention to sign changes and position swaps. In this problem, A = [[0,2],[-1,3]] has det(A) = (0)(3) - (2)(-1) = 0 + 2 = 2, so A^(-1) = (1/2)[[3,-2],[-(-1),0]] = (1/2)[[3,-2],[1,0]]. Choice A is correct because it properly applies the formula with correct signs and positions. Choice B has the wrong sign for -b, choice C incorrectly swaps more elements than required, and choice D has the wrong sign for -c. To help students: memorize the pattern of swapping diagonal elements and negating off-diagonal elements, practice verifying inverses by multiplication, and check determinant calculations carefully.

This question tests AP Precalculus skills involving matrices, specifically calculating the determinant and recognizing when a matrix is not invertible. The determinant of a 2x2 matrix determines whether the matrix is invertible, with det = 0 meaning the matrix is singular (not invertible). In this problem, the matrix A = [[1,-2],[3,-6]] has det(A) = (1)(-6) - (-2)(3) = -6 + 6 = 0. Choice B is correct because the determinant equals 0, which means A is not invertible and cannot be used to solve systems uniquely. Choices A and C represent calculation errors, while choice D (-3) might come from dividing one row by another. To help students: identify proportional rows (row 2 = 3×row 1), recognize that proportional rows always yield det = 0, and understand the connection to linear dependence.

This question tests AP Precalculus skills involving matrices, specifically identifying errors in determinant calculations. The correct determinant formula for a 2×2 matrix [a, b; c, d] is ad - bc, requiring subtraction of the off-diagonal product from the main diagonal product. In this problem, the student incorrectly calculated det([2, 3; 1, 4]) as 2×4 + 3×1 = 11, using addition instead of subtraction. Choice A is correct because the error is using ad + bc instead of ad - bc; the correct calculation should be 2×4 - 3×1 = 8 - 3 = 5. Choice B about swapping rows is incorrect as that would change the matrix entirely, not just the operation used. To help students: use visual aids showing the diagonal products with subtraction signs, create mnemonics like 'main minus off', and practice identifying common calculation errors in peer work.

This question tests AP Precalculus skills involving matrices, specifically calculating inverses using the standard formula for 2×2 matrices. The inverse formula A^(-1) = (1/(ad-bc)) × [d, -b; -c, a] requires first computing the determinant and then applying the adjugate matrix scaled by its reciprocal. In this problem, the matrix A = [5, 1; 2, 1] has determinant det(A) = 5×1 - 1×2 = 5 - 2 = 3, confirming invertibility. Choice A is correct because applying the inverse formula gives A^(-1) = (1/3) × [1, -1; -2, 5] = [1/3, -1/3; -2/3, 5/3]. Choice C incorrectly uses 1/7 as the scalar, suggesting a determinant calculation error of 7 instead of 3. To help students: break down the inverse formula into steps (find det, form adjugate, scale), practice verifying inverses by multiplication, and emphasize sign patterns in the adjugate matrix.