Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations possesses infinitely many solutions when the determinant of its coefficient matrix equals zero, provided the system remains consistent, which requires the determinants of the augmented matrices also to evaluate to zero. The determinant of the coefficient matrix must equal zero for the system to have infinitely many solutions, so we set

. Expanding this along the first row provides $11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24) = 0$, which simplifies to $11(-22) + 118 - 62\lambda = 0$. Solving the expression $-242 + 118 - 62\lambda = 0$ leads to $-124 = 62\lambda$, from which $\lambda = -2$.

Enforcing consistency by setting the determinant of the matrix formed by replacing the first column with the constants -5, 3, and $\mu$ to zero results in

. Expanding this gives $-5(-117 + 95) - 1(-117 - 5\mu) - 2(-57 - 3\mu) = 0$. Simplifying leads to $110 + 117 + 5\mu + 114 + 6\mu = 0$, which combines to $11\mu + 341 = 0$, confirming $\mu = -31$. Substituting these values into the expression $\lambda^4 - \mu$ results in $(-2)^4 - (-31)$, which is $16 + 31$, equaling 47.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, which for a $2 \times 2$ matrix $A$ is given by $A^2 - \text{Tr}(A)A + |A|I = 0$. Given that the sum of the diagonal elements is $\text{Tr}(A) = -3$ and the determinant is $|A| = 2$, substituting these values yields the equation $A^2 + 3A + 2I = 0$. Comparing this result with the provided condition $A^2 + xA + yI = 0$ confirms that the parameters are $x = 3$ and $y = 2$.

Relating these parameters to the geometric properties of the hyperbola yields an eccentricity $e$ such that $e^2 = 3$ and a latus rectum length $\ell = 4$. Consequently, the expression $e^4 + \ell^4$ becomes $3^2 + 4^2$, which evaluates to 25.

Given a symmetric matrix

, the equation

provides the system of equations $a + b = 3$ and $b + d = 7$. Expressing $a$ and $d$ in terms of $b$ as $a = 3 - b$ and $d = 7 - b$ allows the determinant condition $ad - b^2 = 1$ to be written as $(3 - b)(7 - b) - b^2 = 1$. Solving $21 - 10b = 1$ results in $b = 2$, which further determines $a = 1$ and $d = 5$. Therefore, the matrix is

.

Calculating the inverse of this matrix provides

. The condition $A^{-1} = \alpha A + \beta I$ translates to

. Comparing the corresponding entries reveals $2\alpha = -2$, which gives $\alpha = -1$, and $\alpha + \beta = 5$. Consequently, the value of $\alpha + \beta$ is $5$.

To find the determinant of a matrix product, we rely on the fundamental property that the determinant of a product is equal to the product of the determinants of the individual matrices. Additionally, the determinant of a matrix remains unchanged when transposed, meaning $\det(A^T) = \det(A)$. We first compute the determinants for the base matrices: $\det(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3$ and $\det(B) = (1)(1) - (0)(1) = 1$.

For matrix $C = A B A^T$, the determinant is the product of the individual components, $\det(C) = \det(A) \cdot \det(B) \cdot \det(A^T) = 3 \cdot 1 \cdot 3 = 9$. Calculating the determinant for $X = A^T C^2 A$ follows the same multiplicative rule, leading to $\det(X) = \det(A^T) \cdot \det(C^2) \cdot \det(A)$. Substituting the known values, we obtain $\det(X) = 3 \cdot (9^2) \cdot 3$, which results in $9 \cdot 81 = 729$.

Evaluating the determinant of a complex expression involving matrices relies on the fundamental properties $\det(kA) = k^n \det(A)$ and $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$, where $n=3$ for this matrix. Starting from the innermost term, $\det((2A)^{-1})$ is the reciprocal of $\det(2A)$, which equals $\frac{1}{2^3 \det(A)} = \frac{1}{8 \cdot 3} = \frac{1}{24}$. Applying the adjugate property, $\det(\operatorname{adj}((2A)^{-1})) = (\frac{1}{24})^2$, and scaling by 3 yields $\det(3 \operatorname{adj}((2A)^{-1})) = 3^3 \cdot \frac{1}{24^2} = \frac{27}{576} = \frac{3}{2^6}$. Taking the adjugate of this value squares the determinant, and performing the subsequent operations of scaling by $-3$ and $-4$ followed by their respective adjugates continues to apply these scaling and power properties. Following these steps systematically, the overall determinant simplifies to $\frac{3^{20}}{2^{36}}$, which is expressed as $2^{-36} \cdot 3^{20}$. Given the form $2^{m}3^{n}$, we identify $m = -36$ and $n = 20$. The final value is $m + |2n| = -36 + |2(20)| = -36 + 40$, which equals $4$.

A homogeneous system of linear equations possesses a non-trivial solution if and only if the determinant of its coefficient matrix is zero. Calculating the determinant of the given system provides:

$\begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0$

Expanding this determinant leads to $-(\cos^2 \alpha + \sin^2 \alpha) + \sqrt{2} \sin \alpha (\cos \alpha + \sin \alpha) + \sqrt{2} \cos \alpha (\sin \alpha - \cos \alpha) = 0$. By applying the trigonometric identities $\sin 2\alpha = 2 \sin \alpha \cos \alpha$ and $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$, the equation simplifies to $-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0$, which rearranges to $\cos 2\alpha - \sin 2\alpha = -\frac{1}{\sqrt{2}}$. Dividing by $\sqrt{2}$ yields the form $\frac{1}{\sqrt{2}} \cos 2\alpha - \frac{1}{\sqrt{2}} \sin 2\alpha = -\frac{1}{2}$, which is equivalent to $\cos \left( 2\alpha + \frac{\pi}{4} \right) = -\frac{1}{2}$. Solving for the general solution $2\alpha + \frac{\pi}{4} = 2 n \pi \pm \frac{2\pi}{3}$ and restricting $\alpha$ to the interval $\left( 0, \frac{\pi}{2} \right)$, we set $n=0$ and select the positive branch to arrive at $\alpha = \frac{5\pi}{24}$.

For a system of linear equations to possess an infinite number of solutions, the equations must be linearly dependent, which requires the determinant of the coefficient matrix to be zero. By calculating the determinant from the coefficients of $x$, $y$, and $z$, we obtain the expression

. Simplifying this evaluates to $3\lambda - 3 - 4\lambda + 8 - 18$, which simplifies to $-\lambda - 13$. Setting this determinant to zero forces $\lambda = -13$.

With $\lambda = -13$, the system must also satisfy a consistency condition where the augmented matrix does not lead to a contradiction. Subtracting twice the first equation from the second results in $y + 5z = 1$, while subtracting four times the first equation from the third yields $-5y - 25z = \mu - 20$. For the system to have infinite solutions, these equations must represent the same linear relationship, meaning $-5y - 25z$ must be equivalent to $-5(y + 5z)$, or $-5(1)$. Consequently, $\mu - 20 = -5$, which determines that $\mu = 15$. Substituting these values into the required expression, we calculate $\lambda + 2\mu = -13 + 2(15)$, which results in 17.

A system of linear equations possesses infinitely many solutions when the determinant of the coefficient matrix is zero and the determinants of the augmented matrices are likewise zero. Calculating the determinant of the coefficient matrix, defined as

, and setting it to zero results in $1(18-\mu) - 4(14-5\mu) - 1(7-45) = 0$, which simplifies to $\mu = 0$. With this value established, the condition

leads to $\lambda(18) - 4(-6) - 1(6) = 0$, or $18\lambda + 18 = 0$, revealing $\lambda = -1$. Evaluating the requested expression $2\mu + 3\lambda$ gives $2(0) + 3(-1)$, which results in $-3$.

The determinant of a product of matrices is equal to the product of their individual determinants, and for any $3 \times 3$ matrix $M$, the determinant of its adjoint satisfies $|\operatorname{adj}(M)| = |M|^{n-1} = |M|^2$. Given the order $n=3$, $|A|=3$, and $|B|=2$, the expression decomposes into a product of individual determinants: $|A^T| |A| |(\operatorname{adj}(2A))^{-1}| |\operatorname{adj}(4B)| |(\operatorname{adj}(AB))^{-1}| |A| |A^T|$. The term $|A^T| |A| |A| |A^T|$ simplifies to $|A|^4 = 3^4 = 81$.

The internal components are calculated using the property $|\operatorname{adj}(M)^{-1}| = \frac{1}{|\operatorname{adj}(M)|} = \frac{1}{|M|^2}$. Applying this, the term $|(\operatorname{adj}(2A))^{-1}|$ becomes $\frac{1}{|2A|^2} = \frac{1}{(2^3|A|)^2} = \frac{1}{24^2}$, $|\operatorname{adj}(4B)|$ becomes $|4B|^2 = (4^3|B|)^2 = (64 \cdot 2)^2 = 128^2$, and $|(\operatorname{adj}(AB))^{-1}|$ becomes $\frac{1}{|AB|^2} = \frac{1}{(|A||B|)^2} = \frac{1}{6^2}$. Combining these, the total determinant is $81 \cdot \frac{1}{24^2} \cdot 128^2 \cdot \frac{1}{6^2}$, which simplifies to $81 \cdot \frac{16384}{576 \cdot 36}$ and results in a final value of 64.

Determinant properties for a matrix $A$ of order 3 are governed by the relations $\det(kA) = k^3 \det A$, $\det(\operatorname{adj} A) = (\det A)^2$, and $\operatorname{adj}(kA) = k^2 \operatorname{adj} A$. Defining $D = \det A$, the first given expression $\det(3 \operatorname{adj}(2 \operatorname{adj}(D A)))$ can be evaluated by simplifying the inner terms. Since $\operatorname{adj}(D A) = D^2 \operatorname{adj} A$, the inner part becomes $\operatorname{adj}(2 D^2 \operatorname{adj} A)$. Applying the adjoint property, $\operatorname{adj}(2 D^2 \operatorname{adj} A) = (2 D^2)^2 \operatorname{adj}(\operatorname{adj} A) = 4 D^4 (D A) = 4 D^5 A$. The determinant is then $\det(3 \cdot 4 D^5 A) = \det(12 D^5 A) = 12^3 D^{15} \det A = 2^6 \cdot 3^3 \cdot D^{16}$. Matching this with $3^{-13} \cdot 2^{-10}$ results in $D^{16} = 3^{-16} \cdot 2^{-16}$, meaning $D = 2^{-1} \cdot 3^{-1}$.

The second expression $\det(3 \operatorname{adj}(2 A))$ simplifies as $3^3 \det(\operatorname{adj}(2 A)) = 3^3 (\det(2 A))^2$. Substituting $\det(2 A) = 2^3 D = 8 D$, the expression equals $3^3 (8 D)^2 = 3^3 \cdot 64 \cdot D^2 = 3^3 \cdot 2^6 \cdot D^2$. Replacing $D$ with $2^{-1} \cdot 3^{-1}$, the result is $3^3 \cdot 2^6 \cdot (2^{-1} \cdot 3^{-1})^2 = 3^3 \cdot 2^6 \cdot 2^{-2} \cdot 3^{-2} = 3^1 \cdot 2^4$. Given $\det(3 \operatorname{adj}(2 A)) = 2^m \cdot 3^n$, it follows that $m = 4$ and $n = 1$. The required value $|3m + 2n|$ is $|3(4) + 2(1)| = 14$.

The matrix equation

indicates that the sum of the elements in each row of the $3 \times 3$ matrix $A$ is 3. Since all elements in the matrix are non-negative, every individual entry is constrained to the interval $[0, 3]$.

The determinant of a matrix is bounded by the product of the magnitudes of its row vectors. Given that each row sums to 3, the maximum magnitude for a row vector occurs when one element is 3 and the other two are 0, resulting in a magnitude of 3. By multiplying the maximum possible magnitudes for each of the three rows, the determinant is capped at:
$3 \times 3 \times 3 = 27$
This maximum value is attained when the matrix is a diagonal matrix, having 3 at each diagonal position and 0 elsewhere.

The determinant of the adjoint of a $3 \times 3$ matrix $A$ follows the property $\det(\operatorname{adj} A) = |A|^{3-1} = |A|^2$. Iterating the adjoint operator $k$ times for a $3 \times 3$ matrix yields the general result $\det(\operatorname{adj}^{(k)} A) = |A|^{2^k}$. Given $|A|=2$ and $k=2024$, the value is defined as
$n = 2^{2^{2024}}$
To find $n \pmod 9$, we observe the cyclic behavior of the powers of $2$ modulo $9$. Since $2^6 \equiv 1 \pmod 9$, the remainder depends on the exponent $2^{2024}$ modulo $6$. Because $2^{2024}$ is even and $2^{2024} \equiv (-1)^{2024} \equiv 1 \pmod 3$, it must be that $2^{2024} \equiv 4 \pmod 6$, allowing us to write the exponent as $2^{2024} = 6m + 4$ for some integer $m$. Substituting this into the expression for $n$, we obtain $n = 2^{6m+4} = (2^6)^m \cdot 2^4 \equiv 1^m \cdot 16 \equiv 16 \equiv 7 \pmod 9$.

The Cayley-Hamilton theorem establishes that any square matrix satisfies its own characteristic equation, meaning we can derive the matrix properties from $A^3 - 4A^2 + A + 21I = 0$. Since the coefficient of the squared term in the characteristic polynomial corresponds to the negative trace of the matrix, the sum of the diagonal elements $2 + 3 + b$ must equal $4$, which directly yields $b = -1$. The constant term in the characteristic polynomial for a $3 \times 3$ matrix is equal to $-\det(A)$, implying $\det(A) = -21$. Expanding the determinant of matrix $A$ along the first row gives $2(3b - 5) - a(b)$, and substituting $b = -1$ results in $-16 + a$. Equating this to the determinant value $-21$ gives $a = -5$, and evaluating $2a + 3b$ with these values results in $-13$.

The matrix $A = I_2 - 2 MM^T$ exhibits a distinct behavior when squared, as $A^2 = (I_2 - 2 MM^T)(I_2 - 2 MM^T)$ expands into $I_2 - 4 MM^T + 4 MM^T MM^T$. Utilizing the given relation $M^T M = 1$, the term $4 MM^T MM^T$ simplifies to $4 MM^T$, which causes the entire expression to reduce to $A^2 = I_2$. Substituting this into the eigenvalue definition $AX = \lambda X$ yields $A^2 X = \lambda^2 X$, which simplifies to $X = \lambda^2 X$. This implies $\lambda^2 = 1$, meaning the possible values for $\lambda$ are $1$ and $-1$, making the sum of their squares $1 + 1 = 2$.

The condition $A^2 = I$ implies that the square of the determinant of $A$ must be $1$, so $|A|$ equals either $1$ or $-1$, meaning the absolute value $b = | |A| |$ is definitely $1$. By representing $A$ as

, the matrix product $A^2 = I$ necessitates that the off-diagonal entries of

are zero. Since the problem explicitly states that all entries in $A$ are non-zero, $\beta$ and $\gamma$ cannot be zero, which forces the trace of the matrix, $\alpha + \delta$, to be zero, resulting in $a = |\alpha + \delta| = 0$. Substituting these values into the required expression yields $3(0)^2 + 4(1)^2 = 4$.

The behavior of a system of linear equations depends entirely on the determinant of the coefficient matrix, calculated as $\Delta = 7(\alpha + 5)$. A unique solution exists whenever $\Delta \neq 0$, which corresponds to $\alpha \neq -5$. If $\alpha = -5$, the determinant vanishes, and the system becomes either inconsistent or capable of having infinitely many solutions, depending on the consistency of the constant terms.

When $\alpha = -5$, the consistency is determined by the values of $\Delta_1$, $\Delta_2$, and $\Delta_3$, all of which depend on the term $(\beta - 9)$. Specifically, if $\beta = 9$, these determinants are all zero, resulting in infinitely many solutions. If $\beta \neq 9$, at least one determinant is non-zero, making the system inconsistent.

Evaluating the given conditions reveals that for $\alpha = -5$ and $\beta = 8$, the system is inconsistent because $\beta \neq 9$. Conversely, for $\alpha = -5$ and $\beta = 9$, the system correctly produces infinitely many solutions. Furthermore, when $\alpha \neq -5$, the system yields a unique solution, making the claim that the system has infinitely many solutions for $\alpha = -6$ and $\beta = 9$ logically impossible, as $\alpha = -6$ leads to a non-zero determinant.

The determinant of an adjoint matrix follows the fundamental property $|\operatorname{adj} A| = |A|^{n-1}$ for any $n \times n$ matrix $A$. Applying this rule three consecutive times for a $3 \times 3$ matrix results in a determinant of $|A|^{(3-1)^3}$, or $|A|^8$. Since $|A|^8 = 12^4$, the determinant of the original matrix is $|A| = 12^{4/8} = \sqrt{12} = 2\sqrt{3}$.

Determinant properties allow the expression $|A^{-1} \operatorname{adj} A|$ to be calculated as $|A^{-1}| \cdot |\operatorname{adj} A|$. Substituting $|A^{-1}| = \frac{1}{|A|}$ and $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$ into the product yields $\frac{1}{|A|} \cdot |A|^2 = |A|$. Because the determinant of the matrix is $2\sqrt{3}$, the value of the entire expression is $2\sqrt{3}$.

Infinitely many solutions in a system of linear equations occur if and only if the determinant of the coefficient matrix vanishes, combined with the condition that the determinant of the augmented matrix, created by replacing a column with the constant vector, is also zero. Setting the coefficient determinant to zero requires evaluating the following matrix:
$\begin{vmatrix} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{vmatrix} = 0$
The expansion leads to $2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5) = 0$, which simplifies to $50 - 4\lambda + 10 + \lambda - 9 = 0$, or $51 - 3\lambda = 0$, confirming that $\lambda = 17$.

Establishing the value of $\mu$ involves substituting this $\lambda$ into the augmented matrix determinant:
$\begin{vmatrix} 5 & 1 & -1 \\ \mu & -5 & 17 \\ 7 & 2 & -5 \end{vmatrix} = 0$
Expansion along the first row yields $5(25 - 34) - 1(-5\mu - 119) - 1(2\mu + 35) = 0$. Simplifying these terms provides $-45 + 5\mu + 119 - 2\mu - 35 = 0$, which further reduces to $3\mu + 39 = 0$, resulting in $\mu = -13$. Substituting these values into the required expression $(\lambda+\mu)^{2+}(\lambda-\mu)^2$ leads to $(17-13)^2 + (17 - (-13))^2$, which simplifies to $4^2 + 30^2$, yielding 916.

The matrix equation $A^2 + B = A^2 B$ can be transformed by rearranging terms and adding the identity matrix $I$ to both sides, allowing us to factor the expression into $(A^2 - I)(B - I) = I$. Because the product of $(A^2 - I)$ and $(B - I)$ results in the identity matrix, these two matrices must commute, implying that their product in reverse order, $(B - I)(A^2 - I)$, also equals $I$. Expanding both sides of this equality, $B A^2 - B - A^2 + I = A^2 B - A^2 - B + I$, simplifies directly to the commutation property $A^2 B = B A^2$. This fundamental equivalence confirms the relation $A^2 B = (BA)^2$.

Determinant properties and the algebra of adjoint matrices provide the path to solving this problem. For any $m \times m$ matrix $A$, recall that the determinant of a scaled matrix is $|kA| = k^m |A|$, the adjoint scaling property is $\operatorname{adj}(kA) = k^{m-1} \operatorname{adj} A$, and the magnitude of a double adjoint is given by $|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(m-1)^2}$. Since the given matrix $A$ is $3 \times 3$, these rules simplify significantly with $m=3$.

Starting with the expression $|2 \operatorname{adj}(2 \operatorname{adj}(2 A))|$, focus first on the inner adjoint term, where $\operatorname{adj}(2A) = 2^{3-1} \operatorname{adj} A = 4 \operatorname{adj} A$. Replacing this in the original expression yields $|2 \operatorname{adj}(2 \cdot 4 \operatorname{adj} A)|$, which simplifies to $|2 \operatorname{adj}(8 \operatorname{adj} A)|$. Applying the determinant scaling rule for a $3 \times 3$ matrix, this becomes $2^3 |\operatorname{adj}(8 \operatorname{adj} A)|$. Utilizing the identity $|\operatorname{adj} B| = |B|^{m-1}$ for $B = 8 \operatorname{adj} A$, we find $8 \cdot |8 \operatorname{adj} A|^2$. Expanding this further, $|8 \operatorname{adj} A| = 8^3 |\operatorname{adj} A| = 8^3 |A|^2 = (2^3)^3 (2^2) = 2^9 \cdot 2^2 = 2^{11}$. Consequently, the entire expression evaluates to $2^3 \cdot (2^{11})^2 = 2^3 \cdot 2^{22} = 2^{25}$. Setting $2^{25} = 32^n = (2^5)^n = 2^{5n}$, we determine that $5n = 25$, so $n=5$.

To find $\alpha$, compute the determinant of the given matrix $A$ by expanding along the first row: $1(6-1) - 2(2\alpha-1) + 3(\alpha-3) = 2$. This simplifies to $5 - 4\alpha + 2 + 3\alpha - 9 = 2$, which results in $-\alpha - 2 = 2$, identifying $\alpha = -4$. With both values determined, the final calculation $3n + \alpha$ becomes $3(5) + (-4)$, which equals 11.