Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations possesses a non-trivial solution if the determinant of its coefficient matrix is zero, which for this system leads to $14 - 7 \sin 3\theta - 14 \cos 2\theta = 0$, simplifying to $2 - \sin 3\theta - 2 \cos 2\theta = 0$. Substituting the identities $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$ and $\cos 2\theta = 1 - 2 \sin^2 \theta$ yields $4 \sin^3 \theta + 4 \sin^2 \theta - 3 \sin \theta = 0$, which factors into $\sin \theta (2 \sin \theta - 1) (2 \sin \theta + 3) = 0$. Given the potential values $\sin \theta = 0$, $1/2$, and $-3/2$, we observe that only $\sin \theta = 1/2$ provides valid solutions for $\theta$ within the open interval $(0, \pi)$, as $\sin \theta = 0$ occurs only at the interval boundaries and $\sin \theta = -3/2$ is impossible. This identifies exactly two valid values for $\theta$.

Determinant properties dictate that $\det(AB) = \det(A)\det(B)$, $\det(A^T) = \det(A)$, and $\det(A^{-1}) = \frac{1}{\det(A)}$. Applying these to the conditions $\det(ABA^T) = \det(A)^2\det(B) = 8$ and $\frac{\det(A)}{\det(B)} = 8$ creates a system where $\det(A) = 8\det(B)$. Substituting this relationship into the first equation yields $(8\det(B))^2\det(B) = 64\det(B)^3 = 8$, identifying $\det(B) = \frac{1}{2}$ and thus $\det(A) = 4$. Evaluating the target expression $\det(BA^{-1}B^T) = \det(B) \cdot \frac{1}{\det(A)} \cdot \det(B)$ requires calculating $\frac{\det(B)^2}{\det(A)}$. Substituting the determined values results in:

$\frac{(\frac{1}{2})^2}{4} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$

The value of a $3 \times 3$ determinant containing trigonometric functions often simplifies because of the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$, rendering the specific angle irrelevant to the final algebraic result. Expanding $\Delta_1$ along the first row gives $x(-x^{2-}1) - \sin\theta(-x\sin\theta-\cos\theta) + \cos\theta(-\sin\theta+x\cos\theta)$, which simplifies to $-x^3 - x + x(\sin^2\theta + \cos^2\theta)$, ultimately resulting in $\Delta_1 = -x^3$. Repeating this expansion process for $\Delta_2$ yields the same outcome since $x\sin^2 2\theta + x\cos^2 2\theta$ again collapses to $x$, producing $\Delta_2 = -x^3$. Combining these two results, the sum $\Delta_1 + \Delta_2$ equals $-2x^3$.

The matrix $A$ functions as a rotation transformation, and a key property of these matrices is that raising them to an integer power $n$ results in scaling the internal angle by that factor. This leads to the identity
$A^n = \begin{bmatrix} \cos(n\alpha) & -\sin(n\alpha) \\ \sin(n\alpha) & \cos(n\alpha) \end{bmatrix}$
which simplifies the calculation for $A^{32}$ to

.

Aligning this derived matrix with

reveals that $\cos(32\alpha) = 0$ and $\sin(32\alpha) = 1$. These trigonometric values correspond to an angle of $\frac{\pi}{2}$ radians. Therefore, setting $32\alpha$ equal to $\frac{\pi}{2}$ and solving for the angle results in $\alpha = \frac{\pi}{64}$.

The determinant of the matrix

is computed by expanding along the first row, which results in the expression $2((b^{2+}1)(2) - b^2) - b(b(2) - b) + 1(b^2 - (b^{2+}1))$. Simplifying this yields $2(b^{2+}2) - b^2 - 1$, which further reduces to the polynomial $b^2 + 3$. Dividing this determinant by $b$ gives the function $f(b) = b + \frac{3}{b}$ for $b > 0$. To find the minimum value of this sum, we apply the Arithmetic Mean-Geometric Mean inequality, which states that for any positive numbers $x$ and $y$, their sum is at least twice the square root of their product. By setting $x = b$ and $y = \frac{3}{b}$, we find that $b + \frac{3}{b} \geq 2\sqrt{b \cdot \frac{3}{b}}$, leading to a minimum value of $2\sqrt{3}$.

A system of linear equations exhibits infinitely many solutions when the determinant of the coefficient matrix is zero, which indicates that the equations are linearly dependent. Setting the coefficient determinant equal to zero, expressed as:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{vmatrix} = 0$
simplifies to $\alpha - 5 = 0$, confirming that $\alpha$ must be $5$. For the system to maintain consistency under these conditions, the determinant of the matrix formed by replacing the first column with the constant vector must also vanish:
$\begin{vmatrix} 5 & 1 & 1 \\ 9 & 2 & 3 \\ b & 3 & 5 \end{vmatrix} = 0$
Solving this equation leads to $b - 13 = 0$, which identifies $b$ as $13$. Calculating the final value of $b - \alpha$ yields $13 - 5$, resulting in $8$.

Assessing the consistency of a system of linear equations relies on the determinant of the coefficient matrix, $D$. The system is inconsistent when $D$ equals zero while at least one of the modified determinants, such as $D_3$, remains non-zero. Evaluating $D$ for the matrix

results in $a^2 - 3$, which confirms the determinant vanishes when $|a| = \sqrt{3}$. Computing the numerator determinant $D_3$ using the matrix

yields $a - 4$. Since $a - 4$ is non-zero at $|a| = \sqrt{3}$, the equations represent geometric conditions that never share a common intersection, confirming the system is inconsistent for these values.

A system of linear homogeneous equations possesses a non-trivial solution if and only if the determinant of the coefficient matrix is equal to zero. Requiring the determinant of the matrix to be zero yields:
$\begin{vmatrix} 2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1 \end{vmatrix} = 0$
Performing the expansion, $2(k - 2) - 3(1 - (-4)) - 1(-1 - 2k) = 0$, which simplifies to $2k - 4 - 15 + 1 + 2k = 0$, leading to $4k - 18 = 0$ or $k = 9/2$. Solving the system with this value, adding the first equation $2x + 3y - z = 0$ and the third equation $2x - y + z = 0$ results in $4x + 2y = 0$, meaning $x = -y/2$, while subtracting them gives $4y - 2z = 0$, so $z = 2y$. Substituting these relationships into the expression $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ results in $(-1/2) + (1/2) - 4 + 9/2$, which simplifies to $1/2$.

The matrix $P$ can be effectively decomposed into $P = I + X$, where $I$ is the identity matrix and

. By calculating the powers of $X$, it becomes clear that

and $X^3 = 0$. Since the identity matrix commutes with $X$, the binomial expansion allows us to compute $P^5$ as $(I+X)^5 = I + 5X + 10X^2$.

Given the relation $Q = P^5 + I$, we substitute the expanded form to get $Q = 2I + 5X + 10X^2$. Computing this matrix sum yields

. We identify the necessary elements from this matrix as $q_{21} = 15$, $q_{31} = 135$, and $q_{32} = 15$. Applying these values to the required expression gives $\frac{15+135}{15} = \frac{150}{15}$, which results in 10.

This matrix represents a standard two-dimensional rotation, and a known property of this form is that raising it to any power $n$ results in scaling the interior angle, such that

. For a negative exponent like $-50$, the inverse properties of trigonometric functions allow us to express the transformation as:
$A^{-50} = \begin{bmatrix} \cos(50\theta) & \sin(50\theta) \\ -\sin(50\theta) & \cos(50\theta) \end{bmatrix}$
Substituting the given value $\theta = \frac{\pi}{12}$ into the expression $50\theta$ yields $\frac{50\pi}{12}$, which simplifies to $\frac{25\pi}{6}$. Since $\frac{25\pi}{6}$ is coterminal with $\frac{\pi}{6}$ after subtracting two full rotations, we evaluate the trigonometric functions at this reference angle:
$A^{-50} = \begin{bmatrix} \cos\left(\frac{\pi}{6}\right) & \sin\left(\frac{\pi}{6}\right) \\ -\sin\left(\frac{\pi}{6}\right) & \cos\left(\frac{\pi}{6}\right) \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$

Determinants of complex matrices often reveal simple structures through elementary row operations, allowing the expression to be reduced to a manageable algebraic form. Applying the transformation $R_1 \rightarrow R_1 + R_3 - 2R_2$ to the given matrix simplifies the first row to

, which enables the determinant to be calculated as $(d+2)^2 - \sin^2\theta$. Since the value of $\sin^2\theta$ fluctuates between $0$ and $1$, the expression $(d+2)^2 - \sin^2\theta$ attains its minimum when $\sin^2\theta$ is at its maximum of $1$. Setting this minimum value equal to $8$ results in the equation $(d+2)^2 - 1 = 8$, which simplifies to $(d+2)^2 = 9$. Solving for $d$ yields $d+2 = 3$ or $d+2 = -3$, providing $d=1$ or $d=-5$ as the possible values.

For a system of linear equations to possess more than one solution, the equations must be linearly dependent, meaning one equation acts as a combination of the others rather than providing a unique constraint. If we add the first equation, $2x + 2y + 3z = a$, to the third equation, $x - 3y + 2z = c$, we obtain the expression $3x - y + 5z = a + c$. Because this resulting expression perfectly matches the second equation, $3x - y + 5z = b$, the system can only have multiple solutions if the constants satisfy the condition $a + c = b$, which simplifies to $b - c - a = 0$.

An orthogonal matrix satisfies the property $AA^T = I$, which implies that the sum of the squares of the elements in any column must be equal to one. Focusing on the first column of matrix $A$, the sum of the squares of its entries is calculated as:
$0^{2} + p^{2} + p^{2} = 2p^{2}$
Setting this sum equal to the corresponding value of 1 from the identity matrix yields $2p^2 = 1$. Solving this equation gives $p^2 = \frac{1}{2}$, which results in $|p| = \frac{1}{\sqrt{2}}$.

For a system of linear equations to possess infinitely many solutions, the determinant of the coefficient matrix must be zero, and the augmented matrix must maintain a consistent rank lower than the number of variables. Setting the determinant of the coefficient matrix equal to zero,
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0$
simplifies to $2\lambda - 6 - \lambda + 2 + 1 = 0$, which yields $\lambda = 3$. Substituting $\lambda = 3$ into the system gives the equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+3z=\mu$. Subtracting the first equation from the second results in $y+z=1$, and subtracting the second equation from the third results in $y+z=\mu-6$. For the system to be consistent and have infinitely many solutions, these two expressions for $y+z$ must be identical, implying $\mu - 6 = 1$, which leads to $\mu = 7$. Thus, the value of $\lambda + \mu$ is $3 + 7 = 10$.

A homogeneous system of linear equations possesses non-trivial solutions if and only if the determinant of its coefficient matrix is equal to zero, requiring that

. Expanding this determinant yields the cubic equation $2c^3 + 3c^2 - 1 = 0$, which simplifies to the factored form $(c+1)^2(2c - 1) = 0$. Solving this equation identifies the possible values for $c$ as $-1$ and $\frac{1}{2}$, meaning $\frac{1}{2}$ is the greatest value that satisfies the given condition.

Multiplying upper triangular matrices of the form

is simplified by observing that their product results in a matrix of the same form where the top-right element is the sum of the top-right elements of the individual matrices. Given the sequence of matrices from $1$ to $n-1$, the off-diagonal entry of the product is the sum of integers from $1$ up to $n-1$, which is calculated as $\frac{(n-1)n}{2}$. Setting this expression equal to 78 leads to the equation $\frac{(n-1)n}{2} = 78$, or $(n-1)n = 156$, which identifies that $n = 13$. Substituting this value into the matrix

gives

, and its inverse is found by negating the off-diagonal entry, resulting in

.

The roots of the equation $x^{2+}x+1=0$ satisfy the specific properties $\alpha+\beta=-1$ and $\alpha\beta=1$. To evaluate the determinant, replace the first row with the sum of all three rows; since each element in this new row becomes $y+1+\alpha+\beta$, substituting the sum of the roots reduces every entry in the top row to $y$. This allows a factor of $y$ to be extracted from the matrix, leaving a determinant with a leading row of ones. Expanding this resulting matrix naturally leads to a simplified factor of $y^2$, which, when multiplied by the initial $y$, confirms the final result of $y^3$.

The condition $A^{\tau}A = 3I_3$ requires the matrix $A$ to have columns that are mutually orthogonal, with each column having a squared norm of 3. Calculating the diagonal elements of the product $A^{\tau}A$ reveals the squared lengths of these columns, allowing one to establish algebraic relationships between the variables.

Evaluating the first column of $A$, which is

, the squared norm is $0^2 + (2x)^2 + (2x)^2 = 8x^2$. Equating this to 3 leads to $8x^2 = 3$, so $x^2 = \frac{3}{8}$. The second column of $A$ is

, which has a squared norm of $(2y)^2 + y^2 + (-y)^2 = 6y^2$. Equating this to 3 leads to $6y^2 = 3$, so $y^2 = \frac{1}{2}$. The third column

has a squared norm of $1^2 + (-1)^2 + 1^2 = 3$, which is already satisfied.

With two potential values for $x$ and two potential values for $y$, there are $2 \times 2 = 4$ unique pairs of $(x, y)$. Because the squared values $x^2 = 3/8$ and $y^2 = 1/2$ are distinct, the magnitudes of $x$ and $y$ are different, ensuring that $x \neq y$ holds for every combination. This identifies four matrices that satisfy the given criteria.

The most effective approach for evaluating this determinant is to apply row operations that reveal a common factor across all elements in a single row. By adding the second and third rows to the first row, each element in the first row becomes $(a + b + c)$, which can then be factored out of the determinant entirely. With this term removed, column operations such as subtracting the first column from the second and third columns simplify the matrix, eventually reducing the entire determinant to the form $(a + b + c)(a + b + c)^2$.

Comparing this simplified result to the given expression $(a + b + c)(x + a + b + c)^2$, it is clear that $(x + a + b + c)^2$ must equal $(a + b + c)^2$, which implies that $x + a + b + c$ must equal either $(a + b + c)$ or $-(a + b + c)$. Because the condition $x \neq 0$ is specified, we must discard the first possibility, which would lead to $x = 0$. Consequently, the only remaining valid conclusion is that $x + a + b + c = -(a + b + c)$, which simplifies to the result $x = -2(a + b + c)$.

A system of linear equations possesses a unique solution if and only if the determinant of the coefficient matrix is non-zero. Constructing the determinant $\Delta$ from the system's coefficients yields the matrix

. Applying the row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ transforms this into

, which simplifies to $\alpha + \beta + 2$. For a unique solution, we require $\alpha + \beta + 2 \neq 0$, implying $\alpha + \beta \neq -2$. Among the provided options, the pair $(2, 4)$ is the only one satisfying this condition as $2 + 4 = 6$, whereas all other choices result in a sum of $-2$.