Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

Expanding the determinant of the matrix $A$ simplifies the expression to the formula $|A| = 2(1 + \sin^2\theta)$. For $\theta$ \in the range $\left(\frac{3\pi}{4}, \frac{5\pi}{4}\right)$, the sine function $\sin\theta$ varies within the interval $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, which implies that $\sin^2\theta$ satisfies $0 \leq \sin^2\theta < \frac{1}{2}$. By substituting these bounds into the determinant expression, we find that the values of $|A|$ span the interval $[2, 3)$. This specific range is contained within the bounds defined in option B.

Any square matrix can be uniquely represented as the sum of a symmetric component and a skew-symmetric component, where the former is $\frac{1}{2}(M+M^T)$ and the latter is $\frac{1}{2}(M-M^T)$. Given the matrix

, we find its transpose to be

. Applying these decomposition formulas leads to the following matrices:

$A = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix} + \begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} \right) = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix}$

$B = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix} - \begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} \right) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Multiplying these two matrices requires calculating the sum of the products of the respective rows and columns, which results in the final product:

$AB = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -1 & -4 \end{bmatrix}$

Applying the property of determinants, adding the second column to the first column ($C_1 \to C_1 + C_2$) uses the identity $\cos^2\theta + \sin^2\theta = 1$ to simplify the first column entries to $2$, $2$, and $1$, respectively. Subtracting the second row from the first row ($R_1 \to R_1 - R_2$) yields the determinant

Expanding this determinant along the first row results in

, which simplifies to $2(1+4\cos 6\theta) - 4\cos 6\theta = 0$. This equation reduces to $2 + 4\cos 6\theta = 0$, giving $\cos 6\theta = -\frac{1}{2}$. For $\theta \in (0, \frac{\pi}{3})$, the value $6\theta = \frac{2\pi}{3}$ provides the solution $\theta = \frac{\pi}{9}$.

Expanding the determinant along the first row yields the expression $x(-5x^2) + 6(5x - 5) - 1(-5x) = 0$, which simplifies to $-5x^3 + 35x - 30 = 0$. Dividing the entire equation by $-5$ produces the monic cubic polynomial $x^3 - 7x + 6 = 0$. According to Vieta's formulas, for a cubic equation of the form $ax^3 + bx^2 + cx + d = 0$, the sum of the roots is given by $-\frac{b}{a}$. Because the term containing $x^2$ is absent, its coefficient is $0$, making the sum of the real roots $-\frac{0}{1}$, which equals $0$.

A matrix is invertible if and only if its determinant is non-zero, as this is the fundamental condition for the existence of an inverse. The determinant of this matrix simplifies effectively to the expression

. Through the expansion of this determinant and the application of the identity $\cos^2 t + \sin^2 t = 1$, the expression collapses to $5e^{-t}$. Since $e^{-t}$ is strictly positive for all real $t$, the determinant is never zero, which guarantees that the matrix remains invertible for every real number $t$.

A system of linear equations is consistent and possesses infinitely many solutions when the equations are linearly dependent, meaning one can be derived from the others. Adding the first equation $x+y+z=6$ to the third equation $3x+2y-4z=-5$ yields the combined equation $4x+3y-3z=1$. For the system to be dependent, this result must match the second equation $4x + \lambda y - \lambda z = \lambda - 2$. Comparing the coefficients and constant terms on both sides confirms that $\lambda=3$. This value of $\lambda$ is then identified as a root of the quadratic equation $\lambda^2 - 3\lambda - 6 = 0$.

Homogeneous systems of linear equations possess non-trivial solutions only when the determinant of the coefficient matrix is zero, as this indicates the system is singular and allows for non-zero vectors. Calculating the determinant for the provided system leads to the equation $(\lambda-1)^3 = 0$. Solving this polynomial reveals that $\lambda = 1$ is the only value that satisfies the condition. Because there is exactly one specific value for $\lambda$ that results in a non-trivial solution, the set of all such values is a singleton.

The matrix $A$ can be decomposed into $I+N$, where $I$ is the identity matrix and

is a nilpotent matrix satisfying $N^3=0$. Using the binomial expansion for commuting matrices, $A^{20} = (I+N)^{20} = I + 20N + \binom{20}{2}N^2$. Computing the powers of $N$, we find

. Substituting the binomial coefficient $\binom{20}{2} = 190$, the expression simplifies to $A^{20} = I + 20N + 190N^2$. The first column of $A^{20}$ is obtained by multiplying this sum by the first standard basis vector $e_1 = [1, 0, 0]^T$:

$(I + 20N + 190N^2) \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 20 \\ 20 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 190 \end{bmatrix} = \begin{bmatrix} 1 \\ 20 \\ 210 \end{bmatrix}$

Summing the elements of this resulting column gives $1 + 20 + 210 = 231$.

A system of homogeneous linear equations possesses a non-trivial solution if and only if the determinant of the coefficient matrix equals zero, implying that the equations are linearly dependent. We determine the value of $k$ by setting the determinant of the system to zero:

Expanding this determinant yields $1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0$. Simplifying the terms results in $-3k + 8 + 5k + 36 - 6k = 0$, which reduces to $-4k + 44 = 0$, identifying that $k = 11$. Substituting this value back into the original system allows us to express $x$ and $y$ in terms of $z$. By treating $z$ as a parameter $p$, we consider the equations $x + 11y = -3p$ and $3x + 11y = 2p$. Subtracting the first equation from the second yields $2x = 5p$, so $x = \frac{5p}{2}$. Substituting this expression into the first equation provides $11y = -3p - \frac{5p}{2} = -\frac{11p}{2}$, which simplifies to $y = -\frac{p}{2}$. Evaluating the required expression $\frac{xz}{y^2}$ by substituting these values gives $\frac{(\frac{5p}{2})(p)}{(-\frac{p}{2})^2} = \frac{\frac{5p^2}{2}}{\frac{p^2}{4}}$, resulting in a final value of $10$.

The matrix equation $(A-3I)(A-5I)=0$ expands to $A^2 - 8A + 15I = 0$. Since $A$ is non-singular, multiplying this entire equation by $A^{-1}$ yields $A - 8I + 15A^{-1} = 0$, which can be rearranged as $A + 15A^{-1} = 8I$. Multiplying this resulting equation by $\frac{1}{2}$ gives $\frac{1}{2}A + \frac{15}{2}A^{-1} = 4I$. Comparing this to the form $\alpha A + \beta A^{-1} = 4I$, we identify $\alpha = \frac{1}{2}$ and $\beta = \frac{15}{2}$. Consequently, the sum $\alpha + \beta = \frac{1}{2} + \frac{15}{2} = 8$.

A system of linear equations $AX=B$ has no solution if the determinant of the coefficient matrix $A$ is zero and the resulting equations are inconsistent. The coefficient matrix is
$A = \begin{pmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{pmatrix}$
Calculating the determinant gives $\Delta = 1(6 - 10) - a(3 - 2) + 1(5 - 2) = -4 - a + 3 = -a - 1$. Setting $\Delta = 0$ yields $a = -1$. Substituting $a = -1$ into the system produces the equations $x - y + z = 3$, $x + 2y + 2z = 6$, and $x + 5y + 3z = b$. Subtracting the first equation from the second results in $3y + z = 3$, and subtracting the second from the third results in $3y + z = b - 6$. For the system to be inconsistent, these two expressions must contradict each other, requiring $3 \neq b - 6$, which simplifies to $b \neq 9$.

A system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ has no solution if the coefficients satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Applying this ratio condition to the given system $(k+2)x + 10y = k$ and $kx + (k+3)y = k-1$ results in $\frac{k+2}{k} = \frac{10}{k+3}$, which simplifies to the quadratic equation $k^2 - 5k + 6 = 0$. Factoring the quadratic gives $(k-2)(k-3) = 0$, identifying the potential values $k=2$ and $k=3$. For $k=2$, the ratios become $\frac{4}{2} = \frac{10}{5} = \frac{2}{1}$, which implies $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, meaning the system has infinitely many solutions. For $k=3$, the ratios are $\frac{5}{3} = \frac{10}{6} \neq \frac{3}{2}$, which confirms that the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ is satisfied. Thus, exactly one value of $k$ results in no solution for the system.

The relation

implies

. From the determinant property $|cA| = c^n|A|$ where $n=2$, the equation $|3A| = 108$ gives $9|A| = 108$, resulting in $|A| = 12$. Since

, the determinant follows

, yielding $\frac{k^2}{3} = 12$ and $k^2 = 36$. Consequently,

.

Simplifying this determinant relies on recognizing that adding all three rows together produces a common factor of $5x-4$ across the first row. Factoring this out and performing column operations to subtract the first column from the second and third columns transforms the matrix into

. This process yields the final expression $(5x-4)(x+4)^2$, which directly matches the form $(A+Bx)(x-A)^2$. Equating the terms reveals that $A = -4$ and $B = 5$, resulting in the ordered pair $(-4, 5)$.

A system of linear equations $AX = B$ possesses a unique solution if the determinant of the coefficient matrix $A$ is non-zero, expressed as $|A| \neq 0$. For the given system, the coefficient matrix is defined as
$A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{pmatrix}$
Expanding the determinant $|A|$ along the first row yields
$|A| = 1(k - (-2)) - 1(2k - (-3)) + 1(4 - 3)$
$|A| = k + 2 - 2k - 3 + 1$
$|A| = -k$
Setting the determinant not equal to zero for a unique solution gives $-k \neq 0$, which implies $k \neq 0$. Thus, $S$ is the set of all real numbers excluding zero, denoted as $R - \{0\}$.

A system of linear equations lacks a solution when the coefficient determinant is zero and the equations represent inconsistent, non-intersecting conditions. Evaluating the determinant for the given system yields $-(a-1)^2$, implying that $a$ must be $1$ for the system to be potentially inconsistent, a point where the auxiliary determinants $D_1$, $D_2$, and $D_3$ also vanish. Substituting $a=1$ into the system, the first two equations become identical as $x+y+z=1$, while the third equation simplifies to $x+by+z=0$. This system creates an impossible contradiction where $x+y+z=1$ and $x+y+z=0$ must hold simultaneously if $b=1$. Since this is the only value for $b$ that creates such an inconsistency, the set of values for $b$ is a singleton.

Properties involving the adjoint of a matrix are rooted in the fundamental relation $\operatorname{adj}(A) = \lvert A \rvert A^{-1}$ and the specific identity $\operatorname{adj}(\operatorname{adj}(A)) = \lvert A \rvert^{n-2} A$, where $n$ represents the order of the square matrix. For a $3 \times 3$ matrix, setting $n=3$ results in the identity $\operatorname{adj}(\operatorname{adj}(A)) = \lvert A \rvert A$. Evaluating the options against this identity shows that Option B is a direct statement of the result, while Option C simplifies to $\lvert A \rvert^2 \cdot (\frac{1}{\lvert A \rvert} A) = \lvert A \rvert A$, confirming it is also valid. In contrast, Option D simplifies to $\lvert A \rvert \cdot (\frac{1}{\lvert A \rvert} A) = A$, which fails to include the necessary determinant factor and is therefore not always true.

From the given equations $A+B=2B^T$ and $3A+2B=I_3$, isolate $A$ in the first equation to obtain $A=2B^T-B$. Substituting this expression for $A$ into the second equation yields $3(2B^T-B)+2B=I_3$, which simplifies to $6B^T-B=I_3$. Taking the transpose of this equation results in $6B-B^T=I_3$, providing a system of equations involving $B$ and $B^T$.

Expressing $B^T$ as $6B-I_3$ from the transposed equation and substituting this back into $6B^T-B=I_3$ gives $6(6B-I_3)-B=I_3$. Expanding and simplifying leads to $36B-6I_3-B=I_3$, which simplifies further to $35B=7I_3$, resulting in $B=\frac{1}{5}I_3$. Consequently, $A=2(\frac{1}{5}I_3)-\frac{1}{5}I_3=\frac{1}{5}I_3$. Testing the option $10A+5B$, we find $10(\frac{1}{5}I_3)+5(\frac{1}{5}I_3)=2I_3+I_3=3I_3$.

Evaluating the given determinant yields the condition $\cos^3 x - \sin^3 x = 0$, which simplifies to $\tan x = 1$ and identifies $x = \pi/4$ and $x = 5\pi/4$ as the solutions within $[0, 2\pi]$. Applying the tangent addition identity $\tan(\frac{\pi}{3} + x) = \frac{\tan(\frac{\pi}{3}) + \tan x}{1 - \tan(\frac{\pi}{3})\tan x}$ to these values gives $\frac{\sqrt{3} + 1}{1 - \sqrt{3}}$. Rationalizing this result by multiplying the numerator and denominator by $1 + \sqrt{3}$ produces $\frac{4 + 2\sqrt{3}}{-2}$, leading to a value of $-2 - \sqrt{3}$.

Finding the adjoint of a $2 \times 2$ matrix involves swapping the diagonal elements and negating the signs of the off-diagonal entries, effectively transforming a matrix

into

. The calculation begins by determining the resulting matrix from the expression $3A^{2+}12A$. Squaring $A$ yields

, and multiplying this product by $3$ gives

. Adding this result to $12A$, which is

, produces the matrix

. Performing the adjugate operation by swapping the main diagonal values of $72$ and $51$ and changing the signs of $-63$ and $-84$ results in the final matrix

.