Matrices practice · Mixed difficulty

Option C (1) is correct because $\left|A^{\prime} A^{-1}\right| = 1$ since the determinant of the product of a matrix and its inverse is always 1.

Option C ([\beginarraycc1 & 2015 0 & 1\endarray]$\left[\begin{{array}}{{cc}}1 & 2015 \\ 0 & 1\end{{array}}\right]$) is correct because after simplifying the expression $P^T Q^{2015} P$, it reduces to $A^{2015}$, which can be easily calculated.

Option A (Independent of $\alpha$) is correct because the determinant simplifies to an expression that does not contain alpha.

Option C (An odd multiple of $\pi /2$) is correct because the product AB being a null matrix implies that $\cos(\alpha - \beta) = 0$, which means $\alpha - \beta$ is an odd multiple of $\pi /2$.

The correct answer is A (4) because there are only 4 symmetric matrices with integer entries that satisfy the given condition.

Option A (0) is correct because the determinant is 0 due to proportional rows after simplification.

Option C $\left[ {\begin{array}{cccccccccccccccccccc} {{{a^n}}}&0&0\\ 0&{{{a^n}}}&0\\ 0&0&{{{a^n}}} \end{array}} \right]$ is correct because raising a diagonal matrix to the power of n simply raises each diagonal element to the power of n.

Option C (ax(2a+3x)) is correct because after applying row operations, the determinant simplifies to a(a+x)², and then f(2x) - f(x) = ax(2a+3x).

Option C ($2 \lambda + \mu = 14$) is correct because substituting the values of lambda and mu that result in infinitely many solutions into the equation yields a true statement.

Option A (I) is correct because when the given matrix A is squared, the result is the identity matrix I.

Option D (0) is correct because the determinant of the matrix ${A^3} + {B^3}$ is equal to 0, which is derived by manipulating the given matrix equations.

Option C is correct because it results from correctly isolating X and performing the matrix operations.

Option D (None of these) is correct because the inverse matrix calculation results in a different expression than the given options.

Option B (3) is correct. The condition for non-trivial solution implies that the determinant of the coefficient matrix is zero, leading to the relation $\alpha = \beta = \gamma$, and thus $\frac{a^2}{b} = 3$.

Option D (2) is correct because expanding the determinant and solving for x yields x = 2.

Option D (${\left( {a + b + c} \right)^3}$) is correct because the determinant simplifies to this expression after row and column operations.

Option D ($x=0, y=0$) is correct because the determinant simplifies to zero, meaning x and y are both zero.

Option B is correct. The inverse of the matrix is calculated using the formula A^-1 = (1/det(A)) * adj(A).

Option B (A skew-symmetric matrix) is correct because the expression simplifies to a sum of odd powers of a skew-symmetric matrix, which is also skew-symmetric.

Option D (5) is correct because the matrix represents a rotation by $\frac{2\pi}{5}$, and it takes 5 such rotations to return to the identity matrix.

Option A ($\mathrm{A}^{2}+\mathrm{I}=\mathrm{A}(\mathrm{A}^{2}-\mathrm{I})$) is not correct because it simplifies to $0 = -2A$, which is false.

The correct answer is C ($x = \frac{{1}}{{3}}\left( {a + b + c} \right)$) because substituting this value into the determinant makes it equal to zero.

Option C (1,2) is correct because these are the values of $\alpha$ for which the system of equations is consistent, obtained by setting the determinant of the augmented matrix to zero.

Option D (2) is correct because setting the determinant of the matrix to zero (since it's singular) and solving for $\lambda$ gives $\lambda = 2$.

Option D ($\left[\begin{array}{cc}1 & 2500 \\ 0 & 1\end{array}\right]$) is correct because the product of the matrices results in a matrix with the sum of odd numbers from 1 to 99 in the top right element, which equals 2500.

The correct answer is D (0) because when x=0, the determinant becomes a skew-symmetric determinant of odd order, which is always zero.

Option D (more than three) is correct because solving the determinant equation leads to four distinct values of θ in the interval [0, π].

Option D (None of these) is correct because the given options do not match the calculated value of X^n for n=2.

Option D ($\lambda = 3,\mu \ne 10$) is correct because it satisfies the condition for no solution, where the determinant of the coefficient matrix is zero, but at least one of the determinants formed by replacing a column with the constants is non-zero.

Option A ((-6,11)) is correct because it satisfies the equation $6A^{-1} = A^2 + cA + dI$ when solving for c and d.