Matrices practice · Mixed difficulty

Option C $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ is correct because $i^{4n} = 1$ for all integers n, resulting in the identity matrix.

Option C (${\lambda ^2} - \lambda - 6 = 0$) is correct because by setting the determinant of the coefficient matrix to zero and solving, we find that $\lambda = 3$, which satisfies the quadratic equation ${\lambda ^2} - \lambda - 6 = 0$.

Option C (exactly three values of $\lambda$) is correct because the determinant of the coefficient matrix equals zero for $\lambda = 0, 1, -1$, giving three values.

Option C [ \beginarray20c 1&4 0& - 1 \endarray ]$\left[ {\begin{{array}}{{{20}{c}}} 1&4\\\\0&{ - 1} \end{{array}}} \right]$ is correct because it is the matrix A that satisfies the given matrix equation after pre-multiplying by the inverse of the coefficient matrix.

Option B ((|A|+k)^n) is correct because AB = |A|I_n, and substituting this into the expression and using determinant properties leads to the correct result.

Option C ($\sqrt{5}$) is the correct answer because after simplifying the determinant and using trigonometric identities, the maximum value is found to be $\sqrt{5}$.

The correct answer is A (A non-identity symmetric matrix) because after calculating A², M and N can be expressed in terms of I and A, and MN² simplifies to a scalar multiple of the identity matrix, which is symmetric.

Option C (a single ton) is correct because the system has no solution only when $a=1$ and $b=1$, making the set S contain only one element.

The correct answer is skew symmetric because the sum of odd powers of skew-symmetric matrices is also skew-symmetric.

Option C ($A-6 I$) is correct because by Cayley-Hamilton theorem, $A$ satisfies its characteristic equation, which can be used to find $10A^{-1}$.

Option A ([ \beginarray*20c 5&1& - 3 3&2&6 14&5&0 \endarray ]$\left[ {\begin{{array}}{{{*{20}{c}}}} 5&1&{ - 3}\\ 3&2&6\\ {14}&5&0 \end{{array}}} \right]$) is correct because it is the result of the matrix multiplication of A and B.

Option A (-1, -1) is correct because equating the corresponding elements of the matrices leads to e^(x+1) = 1 and e^(y+1) = 1, which gives x = -1 and y = -1.

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

Option D (1) is correct because after calculating the adj(B) and |C|, the ratio |adj(B)|/|C| simplifies to 1.

Option A (Null matrix) is correct because A³ = I, implying A² = A^-1, thus A^-1 - A² = 0.

Option 1 (I) is correct because expanding (I-A)³ and simplifying using A² = A leads to the identity matrix.

Option A (Zero) is correct because the diagonal elements of a skew-symmetric matrix are always zero.

Option B is correct because the adjugate of a 2x2 matrix involves swapping the diagonal elements and negating the off-diagonal elements.

Option D (None of these) is correct because the result depends on whether n is even or odd. When n is odd, A^n is skew-symmetric, and when n is even, A^n is symmetric.

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

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

Option A (0) is correct. By subtracting the given equations and factoring, it's shown that the determinant of (P² + Q²) must be 0 since P is not equal to Q.

Option C (0, 35) is correct because these are the values of x that make the determinant equal to zero.

Option B is correct. Squaring the given matrix and equating it to the identity matrix leads to the relation $1-\alpha^2-\beta \gamma=0$.

Option A ([ \beginarray*20c 5&1& - 3 3&2&6 14&5&0 \endarray ]$\left[ {\begin{{array}}{*{20}{c}} 5&1&{ - 3}\\ 3&2&6\\ {14}&5&0 \end{{array}}} \right]$) is correct because it is the result of the matrix multiplication AB.

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.

Option B ($-\left(3 A^2+2 A+5 I\right)$) is correct because it is derived by multiplying the given equation by $A^{-1}$ and rearranging to solve for $A^{-1}$.

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.