Matrices practice · Mixed difficulty

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 D ($\beta=-8$) is correct because by repeatedly substituting $A^2 = 3A + \alpha I$ into higher powers of A, we can express $A^4$ in terms of A and I, and then compare coefficients.

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 B (${2^n} - 1$) is correct because by expanding $(I+A)^n$ using the binomial theorem and the property $A^2 = A$, we find that the coefficient of A is ${2^n} - 1$.

Option D (${e^{{\Delta _1}/{\Delta _3}}}$ and ${e^{{\Delta _2}/{\Delta _3}}}$) is correct because using Cramer's rule after taking logarithms yields these expressions for x and y.

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 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 A (1) is correct because, given B² = 0, the expression simplifies to the determinant of the identity 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.

The correct answer is C (Has infinitely many non-trivial solutions for $x=2z$) because the determinant of the coefficient matrix is zero, indicating non-trivial solutions, and the relationship $x=2z$ can be derived from the equations.

Option B (Invertible) is correct because the given condition and subsequent algebraic manipulation show that A+I has an inverse.

Option B ($x=\frac{1}{3}(a+b+c)$) is correct because the determinant vanishes when $x=\frac{1}{3}(a+b+c)$.

Option C ($m=n$) is correct because a square matrix is defined as a matrix with an equal number of rows and columns.

Option B (103) is correct because after calculating P⁵⁰ and subtracting I, the ratio (q31 + q32) / q21 equals 103.

The correct answer is exactly one negative value of $\lambda$ because the system is consistent only when $\lambda = -2/3$, which is a single negative value.

Option A ($\lambda = 5$) is correct because when $\lambda = 5$, the determinant of the coefficient matrix is zero, leading to infinitely many solutions for the homogeneous system.

Option A (O) is correct because the expression simplifies to the zero matrix after algebraic manipulation.

Option D (None of these) is correct because the calculated value is 0, which is not among the given options.

Option A (- 12[ \beginarray*20c 4& - 2 - 3&1 \endarray ]$- \frac{1}{2}\left[ {\begin{{array}}{{{*{20}{c}}}} 4&{ - 2}\\ { - 3}&1 \end{{array}}} \right]$) is correct because it is the correct inverse matrix calculated using the formula for a 2x2 matrix inverse.

Option C $\left[\begin{array}{lll}8 & 0 & 0 \\0 & 8 & 0 \\0 & 0 & 8\end{array}\right]$ is correct because $A(\operatorname{Adj} A) = |A|I$, and in this case, $|A| = 8$.

Option D (none of these) is correct. The determinant simplifies to 0, which is not equal to any of the given options.

Option B ($|A B|=0$) is correct because the determinant of a skew-symmetric matrix of odd order is always 0, and $|A B|=|A||B|$. Since $|B|=0$, $|A B|=0$.

Option D (any odd integer) is correct because the equation simplifies to $2\left| {\begin{array}{cccccccccccccccccccc} a&{a + 1}&{a - 1} \\ { - b}&{b + 1}&{b - 1} \\ c&{c - 1}&{c + 1} \end{array}} \right| = 0$ when n is odd.

Option A ($\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$) is correct because ${A^2}$ equals the identity matrix, and therefore ${A^4}$ also equals the identity matrix.