Matrices practice · Mixed difficulty

Option B is correct because the system has no solution when $\lambda = 3$ and $\mu \neq 10$, leading to an inconsistency in the equations.

Option A (-1) is correct because the determinant of the coefficient matrix must be zero for a non-trivial solution, leading to $\lambda = -1$.

Option A ($y(a-b)$) is correct because after applying column and row operations and using the given conditions, the determinant simplifies to $y(a-b)$.

Option D (4 x 3) is correct because A' is a 4x3 matrix, and for A'B and BA' to be defined, B must be a 3x4 matrix, making A'B a 4x4 matrix and BA' a 3x3 matrix. However, the question asks for the type of B, and since A'B is defined, B must have 3 rows, and since BA' is defined, B must have 4 columns.

Option D (None of these) is correct because (A+B)^-1 is a symmetric matrix, but it's not equal to A^-1 + B^-1.

Option C (0) is correct because the trace of a skew-symmetric matrix is always zero, as the diagonal elements of such a matrix are all zero.

Option B is correct because multiplying both sides of the equation AB = 3I by A⁻¹ and simplifying leads to A⁻¹ = (1/3)B.

Option A (Symmetric matrix) is correct because if A is symmetric, then A transpose equals A, and (A^n) transpose also equals A^n.

Option A (0) is correct because the determinant of a skew-symmetric matrix of odd order is always 0, and therefore the determinant of its adjugate is also 0.

Option A ([\beginarraycc1 & -2023 i 0 & 1\endarray]$\left[\begin{{array}}{{cc}}1 & -2023 i \\ 0 & 1\end{{array}}\right]$) is correct because after simplifying the expression using the properties of matrices and the given matrices, we find that $AM^{2023}A^T = B^{2023}$, and $B^{2023}$ follows a pattern that leads to the correct answer.

Option D ($\left( {\frac{a}{b}} \right)$ is one of the cube roots of $- 1$) is correct because the determinant equation simplifies to $a^3 + b^3 = 0$, which means $\left( {\frac{a}{b}} \right)^3 = -1$.

The correct answer is B. By calculating f(A) and adding the given matrix, the result is a zero matrix.

Option A (O) is correct because when A is squared, all elements become zero, resulting in a null matrix.

Option A ($i$) is correct because after applying row operations and substituting the given values, the determinant simplifies to $i$.

Option D ($\frac{{ - 7}}{2}$) is correct because when $\alpha = -\frac{7}{2}$, the determinant of the coefficient matrix is zero, and the system becomes inconsistent.

Option 0 (O) is correct because multiplying (A-2I) and (A-3I) results in a zero matrix.

Option C ($\frac{{ - 1}}{2}$) is correct because after applying row and column operations, the equation simplifies to $\sin 4\theta = - \frac{1}{2}$.

Option A ($\pm 1$) is correct because these are the values of x that make the determinant of the matrix A equal to zero, thus making it singular.

Option A ($2,2,3,4$) is correct because by comparing the corresponding elements of the matrices, we can solve the system of equations to find x=2, y=2, z=3, and w=4.

Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion. Only square matrices can have inverses.

Option A ($x=1, y=-1$) is correct because these values satisfy the matrix multiplication $AA^{-1} = I$.

Option C (1) is correct because the determinant of the coefficient matrix must be zero for a non-zero solution, which leads to k = 1.

Option A ($\left[\begin{array}{ll}-4 & -5 \\ -6 & -7\end{array}\right]$) is correct because solving the system of matrix equations for B yields this result.

Option B (symmetric) is correct because the transpose of (AB - BA) is equal to (AB - BA), satisfying the definition of a symmetric matrix.

Option D (None of these) is correct because $\operatorname{det}(\operatorname{Adj} P) = (\operatorname{det} P)^{n-1}$, and in this case, it equals $(\operatorname{det} P)^2 = 64$, which is not equal to $P^{27}$ or $P^{17}$ or $P^6$.

Option A (1/7) is correct because it represents the fractional part of the expression after simplifying the determinant and using the binomial theorem.

Option C ($\theta=2 n \pi+\frac{\pi}{3}, n \in Z$) is correct because it satisfies the equation $A^T + A = I_2$ by resulting in $\cos \theta = \frac{1}{2}$.

Option C ($A+B$) is correct because given AB=B and BA=A, we can show that A² = A and B² = B, therefore A² + B² = A + B.

Option D ([\beginarraycc1 & 0 -n & 1\endarray]$\left[\begin{{array}}{{cc}}1 & 0 \\ -n & 1\end{{array}}\right]$) is correct because by simplifying the expression A^T C^n A using the given matrices and properties, it reduces to B^n, which is equal to the given matrix.

Option A is correct. The element P_{22} is found by multiplying the corresponding rows and columns of the matrices.