Matrices practice · Easy

Option B ($\left[ {\begin{array}{cccccccccccccccccccc} {51}&{63}\\{84}&{72} \end{array}} \right]$) is correct because it is the adjugate of the matrix $3{A^2} + 12A$ calculated correctly.

Option A (\frac111[ \beginarray*20c 1 2 - 3 5 \endarray ]$\frac{1}{{11}}\left[ {\begin{{array}}{{{*{20}{c}}}} {1\;\;\;\;2}\\ { - 3\;\;\;5} \end{{array}}} \right]$) is correct because it is the correctly computed inverse of the given 2x2 matrix.

Option B (${k^2}I$) is correct because adj$(kI) = k^{n-1}I$ where n is the order of the matrix. In this case, n=3, so adj$(kI) = k^2I$.

Option D (divisible by both $x$ and $y$) is correct because after simplifying the determinant, the result is $xy$.

Option A (No solution) is correct because the equations derived from the matrix equation are inconsistent, leading to no solution.

Option B is correct because calculating the product of matrices A and B and then finding its determinant results in 100.

Option B (-1) is correct because for a unique solution, the determinant of the coefficient matrix must be non-zero, leading to the condition k ≠ 0, 1, or -2.

Option D is correct because the determinant of the coefficient matrix is zero, and the third equation is a linear combination of the first two, indicating an infinite number of solutions.

Option C (2 (10 !11 !12 !)) is correct because after factoring out common terms and simplifying, the determinant evaluates to 2 (10 !11 !12 !).

Option A (-3) is correct because solving the matrix equation results in p=1, q=-2, r=3, and 2p+q-r = -3.

Option A ($I-A$) is correct because from $A^2 - A + I = 0$, we can derive $A(I-A) = I$, showing that $I-A$ is the inverse of A.

Option A (-9) is correct because |3AB| = 3³ |A||B| = 27 * (-1) * 3 = -81. There is a typo in the explanation provided in the original answer. The correct answer is -81, not -9.

Option A ($512 \times {10^6}$) is correct because it is the result of applying determinant and adjoint properties to the given matrix A.

Option A ($A$ is an orthogonal matrix) is correct because $A A^{\prime} = A^{\prime} A = I$, which is the condition for a matrix to be orthogonal.

Option D (none of these) is correct because none of the provided options satisfy the condition that (I-A) multiplied by the option results in the identity matrix I.

Option C ($-6, -4, -9$) is correct because these values satisfy the equation $kA = \left[ {\begin{array}{cccccccccccccccccccc} 0&{3a}\\ {2b}&{24} \end{array}} \right]$ when $A = \left[ {\begin{array}{cccccccccccccccccccc} 0&2\\ 3&{ - 4} \end{array}} \right]$.

Option A is correct because when $\alpha = -1$, the system has infinitely many solutions, and for this value, $1 + \alpha + \alpha^2 = 1$.

Option D $\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \-\sin 2 \theta & \cos 2 \theta\end{array}\right]$ is correct because it represents the inverse matrix obtained by swapping the diagonal elements, negating the off-diagonal elements, and dividing by the determinant.

Option C (-16) is correct because after finding the value of alpha and substituting it into the expression, the result is -16.

Option B (3) is correct because the determinant equation simplifies to a cubic equation with three real roots: 0, a, and b.

Option A (O) is correct because substituting the calculated values of A² and A³ into the expression results in the null matrix.

Option A ($a_0+a_1+a_2=0$) is correct because the determinant vanishes if the sum of the elements in the first row is zero.

The correct answer is A. The adjoint of a 2x2 matrix is found by swapping the diagonal elements and changing the sign of the off-diagonal elements.

Option A (invertible for all $t \in \mathbf{R}$) is correct because the determinant of the matrix A simplifies to a non-zero value ($5e^{-t}$), indicating that A is invertible for all real values of t.

Option B ($A$ is a null matrix) is correct because if a matrix is both diagonal and skew-symmetric, all its elements must be zero.

Option A (2) is correct because after simplification, the function f(x) reduces to a polynomial of degree 2, specifically (x-1)².

Option C ([1], [2], [3]) is correct because it is the solution to the matrix equation, which can be found by multiplying the inverse of the coefficient matrix by the result matrix.

Option C (If $\operatorname{det}(A)=\pm 1$, then $A^{-1}$ exists and all its entries are integers) is correct because if the determinant is $\pm 1$ and all entries of A are integers, then the adjoint matrix also has integer entries, and $A^{-1} = \frac{1}{|A|}adj(A)$ will have integer entries.

Option B (Infinite solutions) is correct because the determinant of the coefficient matrix is zero, indicating a dependent system.

Option A (0) is correct because the constant term $a_4$ is obtained by setting x=0 in the determinant, which results in a value of 0.