Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The equation $AB = B$ implies that the vector $B$ is an eigenvector of the matrix $A$ corresponding to the eigenvalue $1$. Because $B$ is non-zero, the system $(A - I)B = 0$ possesses a non-trivial solution, which mathematically guarantees that the matrix $A - I$ is non-invertible, meaning its determinant must be zero.

Calculating the determinant of this matrix results in the equation $(a-1)(d-1) - bc = 0$. Expanding the parentheses leads to $ad - a - d + 1 - bc = 0$, which can be rearranged to express the determinant of $A$ as $ad - bc = a + d - 1$. Given that $a+d = 2021$, substituting this value yields $2021 - 1$, which equals $2020$.

A system of linear equations fails to have a solution when the main determinant of the coefficient matrix is zero, while the determinant associated with at least one variable is non-zero. Determining the main determinant involves calculating the value:
$\Delta = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix} = 2(-a-1) - 1(a-1) + 1(1+1) = 1-3a$
Setting this to zero to satisfy the condition for inconsistency results in $a = 1/3$. Because the system is inconsistent, the numerator determinant associated with the variable $z$ must not equal zero to avoid an indeterminate state. This determinant is calculated by replacing the third column of the coefficient matrix with the constants from the equations:
$\Delta_3 = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix} = 2(-b-3) - 1(b-3) + 5(1+1) = 7-3b$
Requiring $\Delta_3 \neq 0$ implies $7-3b \neq 0$, or $b \neq 7/3$. The system therefore has no solution when $a = 1/3$ and $b \neq 7/3$.

The matrix transformation $A = XB$ provides the components $a_1 = \frac{b_1 - b_2}{\sqrt{3}}$ and $a_2 = \frac{b_1 + k b_2}{\sqrt{3}}$. Substituting these expressions into the given identity $a_1^2 + a_2^2 = \frac{2}{3}(b_1^2 + b_2^2)$ and multiplying both sides of the equation by 3 yields $(b_1 - b_2)^2 + (b_1 + k b_2)^2 = 2(b_1^2 + b_2^2)$.

Expanding the squared binomials results in $b_1^2 - 2b_1 b_2 + b_2^2 + b_1^2 + 2k b_1 b_2 + k^2 b_2^2 = 2b_1^2 + 2b_2^2$. By canceling the $2b_1^2$ term from both sides, the equation simplifies to $2k b_1 b_2 - 2b_1 b_2 + k^2 b_2^2 + b_2^2 = 2b_2^2$. Rearranging the remaining terms leads to $2b_1 b_2(k - 1) + b_2^2(k^2 - 1) = 0$, which factors into (k - 1)\[2b_1 b_2 + b_2^2(k + 1)]$ = 0$. Given the constraints on the variables, the bracketed expression is non-zero, leaving$k - 1 = 0$, which confirms that the value of$k$ is 1.

A homogeneous system of linear equations admits a non-trivial solution exclusively when the determinant of the coefficient matrix is equal to zero. The condition for the system $4x+\lambda y+2z=0$, $2x-y+z=0$, and $\mu x+2y+3z=0$ is that the determinant

must equal zero.

Expanding this determinant along the first row results in $4(-3-2) - \lambda(6-\mu) + 2(4+\mu) = 0$. Simplifying this expression gives $-20 - 6\lambda + \lambda\mu + 8 + 2\mu = 0$, which reorders to $\lambda\mu - 6\lambda + 2\mu - 12 = 0$. By factoring, we obtain $(6-\mu)(\lambda+2) = 0$. This confirms a non-trivial solution exists whenever $\mu = 6$ for any real $\lambda$, or whenever $\lambda = -2$ for any real $\mu$.

Solving this determinant problem requires applying properties of matrix determinants and adjoints, namely $\det(kA) = k^n \det(A)$ and $\det(\operatorname{Adj}(M)) = (\det(M))^{n-1}$ for an $n \times n$ matrix. First, constructing the matrix $A$ based on the provided rules yields
$A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$
and calculating its determinant gives $\det(A) = 4$. To simplify $\det(3 \operatorname{Adj}(2 A^{-1}))$, apply the property $\operatorname{Adj}(kA) = k^{n-1} \operatorname{Adj}(A)$ for a $3 \times 3$ matrix, which allows the expression to be rewritten as $\det(3 \cdot 2^2 \operatorname{Adj}(A^{-1}))$, or simply $\det(12 \operatorname{Adj}(A^{-1}))$. Utilizing the determinant property $\det(kM) = k^n \det(M)$, the expression becomes $12^3 \det(\operatorname{Adj}(A^{-1}))$, which is $12^3 (\det(A^{-1}))^2$. Substituting $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{4}$ into the result provides $1728 \cdot \frac{1}{16}$, which simplifies to 108.

The core of this problem lies in the property that for any invertible matrix $P$, the expression $P^{-1}AP - I_3$ can be rewritten as $P^{-1}(A - I_3)P$, since $I_3$ is equivalent to $P^{-1}I_3P$. Applying the property of determinants where the determinant of a product equals the product of individual determinants, $|(P^{-1}AP - I_3)^2|$ simplifies to $|P^{-1}(A - I_3)P|^2$. Given that $|P^{-1}| = \frac{1}{|P|}$, the terms $|P^{-1}|$ and $|P|$ cancel out, reducing the entire expression to $|A - I_3|^2$.

We construct $A - I_3$ by subtracting the identity matrix from $A$, resulting in the matrix

. Applying the column operation $C_2 \rightarrow C_2 - C_3$ transforms the second column while leaving the determinant unchanged, yielding the determinant

. Expanding this determinant along the third row gives $-\omega ((1)(-\omega - 2) - (-1)(7 - \omega^2))$, which simplifies to $-\omega(-\omega - 2 + 7 - \omega^2)$. Utilizing the identity $1 + \omega + \omega^2 = 0$, we substitute $-\omega - \omega^2 = 1$ into the expression to find that the determinant equals $-\omega(5 + 1) = -6\omega$. Squaring this result gives $36\omega^2$, which confirms that the value of $\alpha$ is 36.

For a system of linear equations to possess infinitely many solutions, the determinant of the coefficient matrix, denoted as $\Delta$, must equal zero, and the determinants formed by replacing each column with the constant terms, $\Delta_1$, $\Delta_2$, and $\Delta_3$, must also vanish. This condition ensures that the equations are linearly dependent and consistent.

We first evaluate the determinant of the coefficient matrix

to find $\beta$. Expanding this determinant gives $2(-\beta + 3) - 1(\beta + 3) - 1(3 + 3) = -2\beta + 6 - \beta - 3 - 6$, which simplifies to $-3\beta - 3$. Setting this expression to zero results in $-3\beta = 3$, or $\beta = -1$.

Next, we determine $\alpha$ by setting $\Delta_1 = 0$, using the determinant

. Expanding this along the first row, we get $3(1 + 3) - 1(-\alpha + 3) - 1(3\alpha + 3) = 12 + \alpha - 3 - 3\alpha - 3$. This simplifies to $-2\alpha + 6 = 0$, which yields $\alpha = 3$.

Having identified $\alpha = 3$ and $\beta = -1$, we compute the requested value $\alpha + \beta - \alpha \beta$. Substituting these values into the expression results in $3 + (-1) - 3(-1) = 3 - 1 + 3$, which equals 5.

A system of linear equations is inconsistent and yields no solution when its algebraic reduction leads to a logical contradiction, such as zero equalling a non-zero constant. By subtracting five times the first equation from the second, we determine that $x=2$, which simplifies the remaining system to the expressions $y+z=4$ and $2y+\lambda z = \mu-2$. To identify the specific condition for inconsistency, we align the coefficient of $y$ by multiplying the first expression by $2$ to obtain $2y+2z=8$. A contradiction occurs when the coefficients of $z$ are identical but the resulting constant terms differ, meaning $\lambda$ must equal $2$ and $\mu-2$ must not equal $8$, which simplifies to $\mu \neq 10$.

Calculating the determinant of a matrix containing the greatest integer function involves simplifying the matrix through row operations to isolate the constant terms. Since the greatest integer property $[x+n] = [x] + n$ holds for any integer $n$, the elements of the matrix $A$ can be expressed in terms of $[x]$. By subtracting the third row from the first row and the third row from the second row, the $[x]$ terms are eliminated from the top rows, transforming the matrix into

.

Expanding the determinant of this simplified matrix along the first row gives $1([x]+4 + [x]+2) - 0 + (-1)(0 - [x])$, which simplifies to $2[x] + 6 + [x] = 3[x] + 6$. Setting this result equal to the provided determinant value of $192$ gives the equation $3[x] + 6 = 192$, which further simplifies to $3[x] = 186$ or $[x] = 62$. The condition $[x] = 62$ defines the set of values for $x$ as the interval $[62, 63]$.

For a system of homogeneous linear equations to yield a non-trivial solution, the determinant of the coefficient matrix must be zero. By expressing the system as a $3 \times 3$ determinant, we can simplify the matrix by adding the second and third columns to the first column. This operation reveals that the sum in each row becomes $1 + (\cos^2 \theta + \sin^2 \theta) + 4 \sin 3\theta$. Given that the Pythagorean identity establishes $\sin^2 \theta + \cos^2 \theta = 1$, each entry in the first column simplifies to $2 + 4 \sin 3\theta$.

Factoring out this common term $2 + 4 \sin 3\theta$ leaves a remaining determinant that evaluates to one, which indicates that the condition for a non-trivial solution reduces to $2 + 4 \sin 3\theta = 0$. Solving for the trigonometric expression yields $\sin 3\theta = -1/2$. Within the interval $\theta \in (0, \pi/2)$, the value of $3\theta$ that satisfies this equation is $7\pi/6$, which results in $\theta = \frac{7\pi}{18}$.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, which allows us to relate higher powers of the matrix to lower-degree polynomials. For the matrix

, calculating the determinant of $A-\lambda I=0$ results in the characteristic equation $\lambda^{2+}\lambda-2K=0$. Substituting the matrix $A$ into this polynomial yields the relation $A^{2+}A-2KI=0$, which implies $A^2=2KI-A$.

The given expression $A(A^{3+}3I)=2I$ is equivalent to $A^4=2I-3A$. Squaring the relation $A^2=2KI-A$ allows us to express $A^4$ as $A^4=(2KI-A)(2KI-A)=4K^2I-4KA+A^2$. Substituting $A^2=2KI-A$ back into this expression produces $A^4=4K^2I-4KA+2KI-A$, which simplifies to $A^4=(4K^{2+}2K)I-(4K+1)A$. Comparing the coefficients of $A$ from this result and the given equation $A^4=2I-3A$ requires that $4K+1=3$. Solving this linear equation gives $4K=2$, resulting in $K=\frac{1}{2}$.

Any square matrix can be uniquely expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$, which are defined by $P = \frac{1}{2}(A + A^T)$ and $Q = \frac{1}{2}(A - A^T)$. Given the matrix

, the skew-symmetric component is

. Setting the determinant of $Q$ equal to 9, we have:

$\det(Q) = 0 - \left(\frac{3-a}{2}\right)\left(\frac{a-3}{2}\right) = \frac{(a-3)^2}{4} = 9$

Solving $(a-3)^2 = 36$ reveals that $a$ can be either $9$ or $-3$. The symmetric matrix $P$ is calculated as

, and its determinant is $\det(P) = -\frac{(3+a)^2}{4}$. For $a = 9$, $\det(P) = -\frac{12^2}{4} = -36$, and for $a = -3$, $\det(P) = -\frac{0^2}{4} = 0$. Adding these potential values gives a sum of $-36$, and the modulus of this sum is 36.

For any $3 \times 3$ real matrix $A$, the operations involving the adjoint function are governed by the properties $\operatorname{Adj}(kA) = k^2 \operatorname{Adj}(A)$, $\operatorname{Adj}(\operatorname{Adj}(A)) = (\det A) A$, and the determinant relation $\det(\operatorname{Adj}(A)) = (\det A)^2$. The expression $2 \operatorname{Adj}(2 \operatorname{Adj}(\operatorname{Adj}(2 A)))$ can be evaluated systematically from the inside out. Since $\operatorname{Adj}(2A) = 4 \operatorname{Adj}(A)$, the middle portion becomes $\operatorname{Adj}(\operatorname{Adj}(2A)) = \operatorname{Adj}(4 \operatorname{Adj}(A))$, which equals $4^2 \operatorname{Adj}(\operatorname{Adj}(A))$ or $16 (\det A) A$. Multiplying by $2$ and applying the adjoint operator again results in $\operatorname{Adj}(32 (\det A) A) = (32 \det A)^2 \operatorname{Adj}(A)$, which simplifies to $1024 (\det A)^2 \operatorname{Adj}(A)$. Combining this with the initial factor of $2$, the entire expression simplifies to $2048 (\det A)^2 \operatorname{Adj}(A)$, or $2^{11} (\det A)^2 \operatorname{Adj}(A)$.

Taking the determinant of this simplified expression requires the scaling rule $\det(kM) = k^3 \det(M)$. This leads to the equation $(2^{11})^3 \cdot ((\det A)^2)^3 \cdot \det(\operatorname{Adj} A) = 2^{41}$, which expands to $2^{33} \cdot (\det A)^6 \cdot (\det A)^2 = 2^{41}$. Simplifying the exponents gives $(\det A)^8 = 2^{41-33} = 2^8$, which implies that $|\det A| = 2$. Consequently, $\det(A^{2}) = (\det A)^2 = 4$.

The behavior of a system of linear equations is governed by the determinant of the coefficient matrix, denoted as $\Delta$. A non-zero determinant implies a unique solution, while a zero determinant indicates that the system is either inconsistent or possesses infinitely many solutions. For the provided system, the determinant is calculated as:
$\Delta = \begin{vmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{vmatrix} = -a^2 + 7a - 12$
Solving the equation $-a^2 + 7a - 12 = 0$ reveals that $\Delta = 0$ when $a=3$ or $a=4$.

Determining the nature of these cases involves evaluating the auxiliary determinant $\Delta_x$ corresponding to the variable $x$:
$\Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -a & 5 \\ 7 & -2 & -a \end{vmatrix} = 15a + 31$
Substituting the critical values into this expression shows that $\Delta_x$ is non-zero at both $a=3$ and $a=4$, as $15(3) + 31 = 76$ and $15(4) + 31 = 91$. Because the primary determinant vanishes while the auxiliary determinant remains non-zero, the system is inconsistent for these two values of $a$, meaning $n(S_1) = 2$. Infinitely many solutions require the simultaneous vanishing of all auxiliary determinants along with $\Delta$, and since this condition cannot be satisfied for any real $a$, there are no such values, resulting in $n(S_2) = 0$.

The value $a^2 + b^2$ corresponds to the determinant of the matrix

, calculated as $a(a) - (-b)(b)$. Leveraging the property that the determinant of a product equals the product of the determinants, and the determinant of an inverse equals the reciprocal of the determinant, the expression $(I_{2}+A)(I_{2}-A)^{-1}$ simplifies to the ratio of the determinant of $I_{2}+A$ to the determinant of $I_{2}-A$.

Determining the components, $I_{2}+A$ equals

and $I_{2}-A$ equals

. Because both matrices possess a determinant of $1 + \tan^2\left(\frac{\theta}{2}\right)$, their quotient is exactly $1$. Consequently, $a^2 + b^2 = 1$, and multiplying this value by $13$ results in $13$.

Applying row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$ transforms the determinant into a form that reveals common factors. The determinant becomes:
$\begin{vmatrix} \sin x-\cos x & \cos x-\sin x & 0 \\ 0 & \sin x-\cos x & \cos x-\sin x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$
By factoring out $(\sin x - \cos x)$ from the first two rows, the expression simplifies to $(\sin x - \cos x)^2 (\sin x + 2\cos x) = 0$. Equating the first term to zero gives $\tan x = 1$, which provides the solution $x = \frac{\pi}{4}$ within the specified interval of $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$. Setting the second term $\sin x + 2\cos x = 0$ implies $\tan x = -2$, which results in a value outside the given range, confirming that there is only one distinct real root.

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must remain consistent, which implies that the determinants obtained by replacing columns with constant terms must also evaluate to zero. Setting the coefficient determinant $\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix} = 3\alpha+6$ to zero results \in $\alpha = -2$. Using this value for $\alpha$, the condition $\Delta_2 = 0$ for the second determinant gives $\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = \beta-7 = 0$ which identifies $\beta = 7$. Combining these values provides the \sum $\alpha+\beta = 5$.

A system of linear equations is consistent and possesses at least one solution if the determinant of its coefficient matrix is non-zero, yielding a unique solution, or if the determinant is zero while the associated determinants of the augmented matrix also vanish. Constructing the coefficient matrix from the system leads to a determinant defined by the expression
$\Delta = -(|\lambda| + 9)$
When $|\lambda| \neq -9$, the non-zero determinant ensures a unique solution exists, meaning the system is consistent for these values of $\lambda$. In the case where the greatest integer value $|\lambda|$ is $-9$, the determinant equals zero; under these conditions, the determinants $\Delta_1$, $\Delta_2$, and $\Delta_3$ also evaluate to zero, confirming that the system possesses infinite solutions. Since the system remains consistent regardless of whether $|\lambda| \neq -9$ or $|\lambda| = -9$, it is solvable for any real value of $\lambda$.

Determining the behavior of matrix $A$ under repeated multiplication starts with the calculation of its square, which yields

. Squaring this result produces

, and repeating the process one more time results in

. The system of linear equations

therefore expands into the two equations $128x - 128y = 8$ and $-128x + 128y = 64$. Simplifying these yields $x - y = \frac{1}{16}$ and $x - y = -\frac{1}{2}$. Because $x - y$ cannot be both $\frac{1}{16}$ and $-\frac{1}{2}$ at the same time, the two lines described by the system are parallel and distinct, meaning there is no solution.

For a system of linear equations represented by $AX = B$, the existence of a unique solution depends on the determinant of the coefficient matrix, $\Delta$. If $\Delta \neq 0$, the system possesses a unique solution, whereas $\Delta = 0$ implies either no solution or infinitely many solutions. The coefficient matrix for this system is:
$A = \begin{pmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{pmatrix}$
Calculating the determinant yields $\Delta = 1(-4 - k^2) - (-2)(4 - 0) = -4 - k^2 + 8 = 4 - k^2$. This determinant is non-zero whenever $k^2 \neq 4$, which means $k \neq 2$ and $k \neq -2$. Consequently, the condition $k \neq 2, k \neq -2$ guarantees a unique solution, validating statement (A).

When $k = 2$, the determinant becomes zero, and the system is not guaranteed a unique solution. To determine if the system is inconsistent, we evaluate $\Delta_x$, which is the determinant of the matrix formed by replacing the first column with the constants vector. Expanding this determinant results in $\Delta_x = -k^2 - 12k - 20$. Substituting $k = 2$ into this expression gives $\Delta_x = -(2)^2 - 12(2) - 20 = -4 - 24 - 20 = -48$. Since $\Delta = 0$ but $\Delta_x \neq 0$, the system is inconsistent and has no solution when $k = 2$, which confirms statement (D). Because statements (B), (C), and (E) contradict these findings regarding the system's behavior at these specific values of $k$, they are incorrect.