Matrices practice · Medium

Option B is correct. After calculating A^T and A^(-1), and then multiplying them, the resulting matrix is [[cos(2x), -sin(2x)], [sin(2x), cos(2x)]].

Option D (1) is correct because, given M²=O, the expression simplifies to det(I), which equals 1.

The correct answer is D ($a^3$) because by applying column and row operations and using Vieta's formulas, the determinant simplifies to $a^3$.

Option D is correct because the condition for infinitely many solutions is not satisfied for every point on the line $x - 2y + 7 = 0$ except (7,7).

Option C ($2\left( {10!11!12!} \right)$) is correct because after factoring out the factorials and simplifying the determinant, the result is $2\left( {10!11!12!} \right)$.

Option C (${I}+{P}$) is correct because after performing the matrix multiplication and applying trigonometric identities, the result simplifies to ${I}+{P}$.

Option D ([\begincc36 & -32 0 & 4\endcc]$\left[\begin{{cc}}36 & -32 \\ 0 & 4\end{{cc}}\right]$) is correct. By finding the matrix A such that A multiplied by the given matrix results in a scalar matrix, and using the determinant condition, we can find A².

Option C (10) is correct because the maximum value of the determinant is found by simplifying the expression and maximizing the resulting trigonometric function.

Option B (-42, -3, 42, -3) is correct because it accurately calculates the minors and cofactors of the specified elements in the matrix.

Option C (0) is correct because by simplifying the determinant using properties of cube roots of unity, the determinant evaluates to 0.

Option C ([ \beginarray*20c\frac\sqrt 3 2&12 - 12&\frac\sqrt 3 2\endarray ]$\left[ {\begin{{array}}{{{*{20}{c}}}}{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\{- \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}\end{{array}}} \right]$) is correct because it is the result of calculating ${A^{ - 50}}$ using the given matrix A and $\theta = \frac{\pi }{{12}}$.

Option C ($\frac{{-2}}{{3}}$) is correct because by substituting the idempotent matrix property into the given equation and simplifying, we can solve for $\alpha$.

Option D (None of these) is correct because the determinant evaluates to zero regardless of the values of a, b, and c.

Option A (Independent of $\alpha$) is correct because after applying trigonometric identities and simplifying, the determinant's value only depends on $\beta$.

Option B ($-\cos 2 \theta, \sin 2 \theta$) is correct because the matrix multiplication and simplification using trigonometric identities lead to these values for $a$ and $b$.

Option C (-3) is correct because by simplifying the determinant and comparing coefficients, B+C is found to be -3.

Option A (4A - 3I) is correct because when A is squared, it results in 4A - 3I, which can be verified through matrix multiplication and simplification.

The correct answer is B (2) because the determinant of the coefficient matrix must be non-zero for a unique solution, leading to $\lambda = \pm 1$, and $a^2 + b^2 = 1 + 1 = 2$.

Option A ($\left[ {\begin{array}{cccccccccccccccccccc}1&0\\{25}&1\end{array}} \right]$) is correct because the pattern of the matrix power leads to the correct element in the lower left corner after 50 iterations.

Option D ($A$ is independent of $\theta$) is correct because after applying row operations and simplifying, the determinant's value does not depend on $\theta$.

Option B (\beginarrayll0 & i i & 0\endarray)$\left(\begin{{array}}{{ll}}0 & i \\ i & 0\end{{array}}\right)$ is correct because after calculating the powers of matrix M, the series simplifies to M itself.

Option B ($(\operatorname{adj} B)(\operatorname{adj} A)$) is correct because the adjoint of a product of matrices is the product of their adjoints in reverse order.

Option D (2) is correct. The value of the expression is found by manipulating the determinant and simplifying the resulting equation.

Option C (6) is correct because solving the determinant equation yields the value of lambda, and substituting it into the expression gives 6.

Option B (Divisible by both $x$ and $y$) is correct because after applying column operations, the determinant simplifies to $xy$, which is divisible by both $x$ and $y$.

Option A (0, 3) is correct because solving the determinant equation gives the values of y as 0 and 3.

Option A ($\pm \frac{{1}}{{5}}$) is correct. Using the property $|k \cdot adj A| = k^n |A|^{n-1}$ for a $n \times n$ matrix and solving for $|A|$ gives the answer.

The correct answer is Null matrix because when $\theta$ and $\phi$ differ by an odd multiple of $\frac{\pi}{2}$, $E(\theta)E(\phi)$ results in a zero matrix.

Option A (0) is incorrect. Option B (27) is incorrect. Option C (81) is incorrect. Option D (162) is correct because it uses properties of determinants and arithmetic progressions to simplify the expression and find the value.

Option B ($abc$) is correct because by setting the determinant of the coefficient matrix to zero and simplifying, we can show that $ab + bc + ca = abc$.