Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The determinant of a product is the product of the determinants, and the determinant of an inverse is the reciprocal of the determinant. This property simplifies the expression $\left| P^{-1} A P - 2 I \right|$, which can be rewritten as $\left| P^{-1} (A - 2 I) P \right|$. Because $\left| P^{-1} \right| \cdot \left| P \right| = 1$, the terms cancel out, leaving only the determinant of $A - 2 I$.

Substituting the given matrix $A$ and subtracting $2 I$ results in the matrix

. Computing the determinant by expanding along the first row provides $0(0 - 33) - 1(0 - 33) + 2(18 - 0)$, which simplifies to $33 + 36 = 69$. Since 69 can be factored into the prime numbers 3 and 23, adding these values gives a sum of 26.

Computing the quadratic form $A' B A$ where

consists entirely of ones is equivalent to summing all the elements within the matrix $B$, because multiplying a matrix by a column vector of ones effectively calculates the sum of its rows, and subsequent multiplication by a row vector of ones totals those sums. Evaluating the row vector

aggregates the columns of $B$ into the components $[(9^2 + 12^2 - 15^2), (-10^2 + 13^2 + 16^2), (11^2 - 14^2 + 17^2)]$. Substituting the numerical squares yields $[(81 + 144 - 225), (-100 + 169 + 256), (121 - 196 + 289)]$, which simplifies to the vector $[0, 325, 214]$. Finishing the multiplication $A' B A$ by applying the final column vector $A$ adds these components together, resulting in $0 + 325 + 214 = 539$.

Matrix conjugation through an orthogonal matrix simplifies power calculations because such matrices satisfy $P^T P = I$. When we substitute the expression $Q = PAP^T$ into the term $P^T Q^{2007} P$, the repeated products of the middle terms collapse in a telescopic fashion. Specifically, the interior instances of $P^T P$ reduce to the identity matrix, effectively simplifying the entire expression to $(P^T P) A^{2007} (P^T P)$, which is simply $A^{2007}$.

For the matrix

, raising it to the power $n$ yields the general form

. Thus, for $n = 2007$, we obtain

. By mapping this to the matrix

, we identify $a = 1$, $b = 2007$, $c = 0$, and $d = 1$. Substituting these values into the final expression $2 a + b - 3 c - 4 d$ gives $2(1) + 2007 - 3(0) - 4(1)$, which simplifies to $2 + 2007 - 4$, resulting in 2005.

The property that the determinant of an adjoint matrix is related to the determinant of the original matrix by the formula $|\text{adj}(A)| = |A|^{n-1}$ provides the foundation for solving this problem. For a $3 \times 3$ matrix with $|A|=2$, the determinant of the given adjoint matrix $B$ is $|B| = 2^{3-1} = 4$. Expanding the determinant of

along the first row involves computing $1(8-3\alpha) - 3(4-3\alpha) + \alpha(\alpha-2\alpha)$, which simplifies to $-\alpha^2 + 6\alpha - 4$. Setting this equal to the previously determined value of 4 creates the quadratic equation $-\alpha^2 + 6\alpha - 4 = 4$, or $\alpha^2 - 6\alpha + 8 = 0$. Factoring this leads to $(\alpha-4)(\alpha-2) = 0$, yielding potential values $\alpha=4$ or $\alpha=2$. Since the condition $\alpha > 2$ is required, $\alpha=4$ is the only valid solution.

Substituting $\alpha=4$ into the matrix $B$ results in

, and the expression to evaluate becomes

. Multiplying the initial row vector by the matrix $B$ produces the vector

by calculating $4(1)-8(1)+4(4) = 12$, $4(3)-8(2)+4(4) = 12$, and $4(4)-8(3)+4(4) = 8$. Completing the calculation by multiplying this resulting vector with the final column vector

gives $12(4) + 12(-8) + 8(4)$, which evaluates to $48 - 96 + 32 = -16$.

The determinant remains unchanged when adding one column to another, a property that allows one to add the second and third columns to the first to reveal a common factor of $\cos x + 2\sin x$. This operation simplifies the equation to $(\cos x + 2\sin x)(\cos x - \sin x)^2 = 0$, implying that solutions occur where $\cos x = -2\sin x$ or $\cos x = \sin x$, which is equivalent to $\tan x = -\frac{1}{2}$ or $\tan x = 1$. Within the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$, the condition $\tan x = 1$ yields $x = \frac{\pi}{4}$, while $\tan x = -\frac{1}{2}$ yields $x = -\tan^{-1}(\frac{1}{2})$, both of which lie within the specified range. The system therefore has exactly two distinct real roots.

The matrix $A$ follows a cyclic pattern where $A^3 = I$. Because of this periodicity, the expression $B_0 = A^{49} + 2A^{98}$ simplifies by reducing the powers modulo 3 to $B_0 = A^1 + 2A^2$. This yields the matrix
$B_0 = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$
with a determinant of $\det(B_0) = 0(0 - 2) - 1(0 - 1) + 2(4 - 0) = 9$.

For any $3 \times 3$ matrix $M$, the determinant of its adjugate relates to its determinant by the property $\det(\operatorname{adj}(M)) = (\det(M))^2$. Applying this property iteratively for $B_n = \operatorname{adj}(B_{n-1})$ means that each step squares the determinant of the previous matrix, leading to $\det(B_n) = (\det(B_{n-1}))^2$. Extending this to the fourth iteration results in $\det(B_4) = (\det(B_0))^{2^4} = (\det(B_0))^{16}$. Substituting the calculated value of the determinant gives $9^{16} = (3^2)^{16} = 3^{32}$.

The determinant of the adjoint of any square matrix $A$ of order $n$ follows the property $|\operatorname{adj} A| = |A|^{n-1}$, which reduces to $|\operatorname{adj} A| = |A|^2$ for a $3 \times 3$ matrix. Calculating the determinant of the given matrix yields $|A| = 1(1 - 0) + a(0 - (-a)) = 1 + a^2$, and consequently, the adjoint determinant is $|\operatorname{adj} A| = (1 + a^2)^2$.

As $a$ ranges through the set $S = \{\sqrt{n} : 1 \leq n \leq 50, n \text{ is odd}\}$, each term in the summation becomes $((\sqrt{n})^2 + 1)^2$, which simplifies to $(n + 1)^2$. Summing this expression for the 25 odd values of $n$ from 1 to 49 results in the sequence $2^2, 4^2, 6^2, \dots, 50^2$, which can be rewritten as $4 \sum_{k=1}^{25} k^2$. Applying the standard sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ with $n=25$, the total sum is $4 \times \frac{25 \cdot 26 \cdot 51}{6} = 22100$. Given that this sum equals $100\lambda$, it follows that $\lambda = 221$.

Since the product $AA^T$ equals $9I$, the rows of matrix $A$ are orthogonal to each other, meaning the dot product of any two distinct rows must result in zero. Calculating the dot product of the third row with the first and second rows creates a system of linear equations that allows us to find the unknown variables.

For the first and third rows, the dot product is $1(a) + 2(2) + 2(b) = 0$, which simplifies to $a + 2b = -4$. Similarly, the dot product of the second and third rows is $2(a) + 1(2) - 2(b) = 0$, which simplifies to $2a - 2b = -2$. Adding these two resulting equations eliminates the $b$ variable to yield $3a = -6$, confirming that $a = -2$. Substituting this value into the equation $a + 2b = -4$ leads to $-2 + 2b = -4$, which simplifies to $2b = -2$ and identifies $b = -1$. The resulting ordered pair is $(-2, -1)$.

The condition given for the matrix entries $x \sin \theta = y \sin\left(\theta+\frac{2\pi}{3}\right) = z \sin\left(\theta+\frac{4\pi}{3}\right)$ relies on the trigonometric identity $\sin \theta + \sin\left(\theta+\frac{2\pi}{3}\right) + \sin\left(\theta+\frac{4\pi}{3}\right) = 0$. By dividing the given equalities by the product of the sine terms or observing the relationship between the reciprocals, it follows that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$, which is algebraically equivalent to $xy + yz + zx = 0$. If the trace of the matrix, defined as $x + y + z$, were equal to zero, then the identity $(x+y+z)^2 = x^{2+}y^{2+}z^{2+}2(xy+yz+zx) = 0$ would imply that $x^{2+}y^{2+}z^2 = 0$. For real numbers, this necessitates $x=y=z=0$, which contradicts the problem's premise that the entries are non-zero, making the first statement false.

For the second statement, the properties of the adjoint matrix for a $3 \times 3$ matrix $R$ dictate that $\operatorname{adj}(\operatorname{adj}(R)) = |R|^{3-2} R = |R| R$. The trace of this resulting matrix is therefore $|R| \operatorname{trace}(R)$, which expands to $xyz(x+y+z)$. As established by the earlier derivation, the expression $xyz(x+y+z)$ cannot equal zero because the entries are non-zero and their sum cannot be zero. In formal logic, a conditional statement with a false premise-in this case, that the trace is zero-is considered vacuously true regardless of the conclusion. Since the condition required for the trace to be zero is impossible, the statement regarding the implication holds true.

Factoring out $p!$, $(p+1)!$, and $(p+2)!$ from the respective rows of the determinant simplifies the matrix into a structure where the entries involve $1$, $(p+1)$, and $(p+2)$ terms. Applying row operations $R_2 - R_1$ and $R_3 - R_2$ reduces the determinant to $\Delta = 2 \cdot p! \cdot (p+1)! \cdot (p+2)!$. By breaking down the factorials into their components, the expression becomes $\Delta = 2 \cdot (p!)^3 \cdot (p+1)^2 \cdot (p+2)$. The prime $p$ is contained within the $(p!)^3$ term, contributing an exponent of $3$ to the value of $\Delta$, meaning the maximum value of $\alpha$ is $3$. The prime $p+2$ appears as a single linear factor in the expression, indicating that the maximum value of $\beta$ is $1$. Adding these values together yields $3 + 1 = 4$.

This matrix represents a rotation in space, where the inverse of rotating by an angle $x$ is simply rotating by $-x$. When computing the product $f(x)f(-x)$, the resulting matrix is the identity matrix

, confirming that $f(-x)$ is indeed the inverse of $f(x)$.

Multiplying $f(x)$ by $f(y)$ involves standard row-by-column operations that rely on the trigonometric sum identities $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$. Carrying out this matrix multiplication simplifies the entries to the exact structure of $f(x+y)$, meaning both statements are correct.

The equation $\det(A - \alpha I) = 0$ is the characteristic equation used to determine the eigenvalues of matrix $A$. Calculating the determinant of $A - \alpha I$ yields the cubic polynomial expression $\alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0$. This polynomial factors into $(\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0$, which provides the real roots $\alpha = 8$ and $\alpha = -8 \pm 2\sqrt{6}$. Because $p$ represents the largest real root of this equation, it follows that $p = 8$.

Substituting $p = 8$ into the circle equation yields $(x - 8)^2 + (y - 16)^2 = 320$. Setting $y = 0$ to find the intersection points with the x-axis, the equation becomes $(x - 8)^2 + (-16)^2 = 320$, which simplifies to $(x - 8)^2 = 64$ and results in $x - 8 = \pm 8$, identifying the points $(16, 0)$ and $(0, 0)$. Similarly, setting $x = 0$ to determine the intersections with the y-axis produces $(-8)^2 + (y - 16)^2 = 320$, which simplifies to $(y - 16)^2 = 256$ and gives $y - 16 = \pm 16$, identifying the points $(0, 32)$ and $(0, 0)$. Combining these results, the circle intersects the coordinate axes at the three unique points $(16, 0)$, $(0, 32)$, and $(0, 0)$.

The determinant of the adjoint of a matrix $M$ is defined by the property $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$ for an $n \times n$ matrix. Using the provided transformation data, we construct the matrix

.

We express the target matrix as the product $A^2 + A = A(A + I)$. The matrix $A + I$ is calculated as

. Computing the determinant of this sum, we find $\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9) = 9 + 7 - 8 = 8$. Since $\det(A) = 1$, the determinant of the product is $\det(A^2 + A) = \det(A) \times \det(A + I) = 1 \times 8 = 8$. Applying the adjoint property for a $3 \times 3$ matrix, the final value is $8^{3-1} = 8^2 = 64$.

The behavior of the equation $A^n + (\omega B)^n = A + B$ is determined by the matrix powers $A^n$ and $B^n$. Matrix $A$ satisfies the idempotent property $A^2 = A$, which implies $A^n = A$ for any positive integer $n$. Defining $B = A - I$ leads to $B^2 = -B$, establishing a cyclic behavior for powers of $B$ where $B^n = B$ for all odd $n$, and $B^n = -B$ for all even $n$.

Substituting these results into the given equation transforms it into $A + \omega^n B^n = A + B$, which simplifies to $\omega^n B^n = B$. For this equality to hold, we require both $\omega^n = 1$ and $B^n = B$. The condition $\omega^n = 1$ is satisfied only when $n$ is a multiple of 3, while $B^n = B$ is satisfied when $n$ is odd. Combining these two conditions restricts $n$ to odd multiples of 3 within the range 1 to 100.

These values of $n$ are $3, 9, 15, \ldots, 99$. This is an arithmetic progression with the first term $a = 3$, last term $l = 99$, and a common difference $d = 6$. Using the formula $l = a + (k-1)d$, we find $99 = 3 + (k-1)6$, which simplifies to $96 = (k-1)6$, or $16 = k-1$, giving $k = 17$. There are 17 such elements.

The core of this problem lies in applying the standard properties of the adjoint matrix, specifically the identity $\operatorname{adj}(\operatorname{adj}(A)) = |A|^{n-2} A$ and the definition $A \operatorname{adj}(A) = |A|I$, where $|A|$ is the determinant of the matrix $A$ and $n$ represents the order of the matrix. For the matrix

, the determinant is $|A| = -1$, which allows us to simplify the adjoint terms based on the given equation. Given $n=3$, the term $\operatorname{adj}(\operatorname{adj}(A))$ reduces to $|A|A = -A$, and the term $\operatorname{adj}(A)(\operatorname{adj}(\operatorname{adj}(A)))$ simplifies to $\operatorname{adj}(A)(-A) = -|A|I = -(-1)I = I$. Substituting these expressions, the original equation becomes

.

Calculating the square of matrix $A$ yields

. By substituting this into our simplified matrix equation, we obtain

. Comparing the elements in the first row and second column provides the equation $-1 - \alpha(1) = -2$, which gives $\alpha = 1$. Similarly, comparing the elements in the second row and second column results in $1 - \alpha(0) + \beta = 0$, identifying $\beta = -1$. Evaluating the final required value, $(\alpha - \beta)^2$ becomes $(1 - (-1))^2 = 2^2 = 4$.

The $n^{\text{th}}$ term of an arithmetic progression is given by the general formula $T_n = a_1 + (n-1)d$. Given three arithmetic progressions with a common difference $d$ and starting terms $A, A+1,$ and $A+2$, their respective terms are $a = A+6d$, $b = A+1+8d$, and $c = A+2+16d$. Since the problem specifies $a=29$, we establish the linear relation $A+6d=29$. By substituting these expressions into the provided determinant equation and solving the resulting system, we determine the unique values of the constants to be $A=-7$ and $d=6$.

To find the sum of the first $20$ terms of the new arithmetic progression, we first determine its starting term, $a_1 = c-a-b$. Substituting the established expressions for $a, b,$ and $c$ yields $a_1 = (A+16d+2) - (A+6d) - (A+8d+1)$, which simplifies to $-A + 2d + 1$. Plugging in $A=-7$ and $d=6$, we find the starting term $a_1 = -(-7) + 2(6) + 1 = 20$. The common difference for this new progression is $\frac{d}{12} = \frac{6}{12} = 0.5$. Applying the sum formula $S_n = \frac{n}{2}[2a_1 + (n-1)d_{new}]$, we calculate the sum for $n=20$ as $S_{20} = \frac{20}{2}[2(20) + (19)(0.5)]$. This results in $10[40 + 9.5]$, which equals $495$.

Decomposing the matrix $A$ into the sum of the identity matrix $I$ and a strictly lower triangular matrix $P$, defined as the matrix with entries 3 and 9 in the second and third rows respectively, simplifies the task, as $P^3$ becomes the zero matrix. Since $A = I + P$, the expression $A^{99} = (I+P)^{99}$ expands using the binomial theorem into $I + 99P + \binom{99}{2}P^2$, where higher-order terms vanish due to the nilpotency of $P$. Consequently, the matrix $B = A^{99}-I$ simplifies to $99P + \binom{99}{2}P^2$.

Calculating the required entries specifically, the element $b_{21}$ is $297$, while $b_{32}$ is also $297$. The element $b_{31}$ arises from both terms in the expansion, resulting in a value of $44550$. Substituting these into the requested ratio yields $\frac{44550 - 297}{297}$, which simplifies to $149$.

This problem is solved by recognizing that the given determinant can be decomposed into the product of two identical $3 \times 3$ matrices, where one is the transpose of the other. The original matrix is formed by the product

, which makes the determinant equal to the square of

. Applying column transformations, specifically subtracting the first column from the second and third columns, simplifies this determinant to $(\alpha-1)(\beta-1)(\beta-\alpha)$, and squaring this term results in $(\alpha-1)^2(\beta-1)^2(\alpha-\beta)^2$. Since $(1-\alpha)^2 = (\alpha-1)^2$ and $(1-\beta)^2 = (\beta-1)^2$, matching this to the provided expression $K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2$ reveals that $K$ equals $1$.