Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations is inconsistent and lacks a solution when the determinant of the coefficient matrix equals zero, provided that the system's augmented matrix reveals a contradiction. We define the coefficient matrix determinant for the given system as:
$\Delta = \begin{vmatrix} 6\lambda & -3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda \end{vmatrix} = 0$
Expanding this determinant along the first row yields $6\lambda(18\lambda^2 - 8) + 3(6\lambda - 12) + 3(4 - 18\lambda) = 0$. Simplifying this expression results in $108\lambda^3 - 48\lambda + 18\lambda - 36 + 12 - 54\lambda = 0$, which further reduces to $108\lambda^3 - 84\lambda - 24 = 0$. Dividing this cubic equation by 12 gives $9\lambda^3 - 7\lambda - 2 = 0$, which factors into $(\lambda - 1)(3\lambda + 1)(3\lambda + 2) = 0$.

The roots of this equation are $\lambda = 1, -\frac{1}{3}, -\frac{2}{3}$, and testing these values shows that the augmented determinants are non-zero, confirming an inconsistent system for each. Therefore, the set $S$ contains all three of these values. The final task requires computing $12 \sum_{\lambda \in S} |\lambda|$, which corresponds to $12 \times (|1| + |-\frac{1}{3}| + |-\frac{2}{3}|)$. Summing these absolute values gives $1 + \frac{1}{3} + \frac{2}{3} = 2$, and the product of this result with 12 yields 24.

A system of linear equations possesses a unique solution when the determinant of its coefficient matrix, $\Delta$, is non-zero, whereas the possibility of infinitely many solutions is generally contingent upon $\Delta$ being equal to zero. Evaluating the determinant for this system results in $\Delta = a(15 a^{2+}31 a+36)$. Since the quadratic expression $15 a^{2+}31 a+36$ does not produce any real roots, the determinant vanishes exclusively when $a = 0$.

Because the provided domain restricts $a$ to $\mathbb{R}-\{0\}$, the determinant is never zero for any valid input. This behavior ensures the system always admits a unique solution for every $a$ in the specified range, establishing $S_1 = \mathbb{R}-\{0\}$. Consequently, as the condition $\Delta = 0$ is never satisfied within the domain, no values of $a$ exist to yield infinitely many solutions, resulting in $S_2 = \Phi$.

Since the determinant equality holds for any real value of $x$, choosing $x=0$ serves as an efficient strategic step to isolate the parameter $\lambda$. Substituting $x=0$ transforms the matrix into a diagonal one with entries $1$, $\lambda$, and $\lambda^2$, making the determinant simply the product $1 \cdot \lambda \cdot \lambda^2 = \lambda^3$. Evaluating the right-hand side of the equation at $x=0$ yields $\frac{9}{8} \times 81$, which simplifies to $\frac{729}{8}$.

Equating these two results gives $\lambda^3 = \frac{729}{8}$, and taking the cube root of both sides confirms that $\lambda = \frac{9}{2}$. Given the roots of the required equation are $\lambda$ and $\frac{\lambda}{3}$, the values are $\frac{9}{2}$ and $\frac{3}{2}$. Constructing the quadratic equation x^2 - (\text{\sum of roots})x + (\text{product of roots}) = 0, the sum is $\frac{9}{2} + \frac{3}{2} = 6$ and the product is $\frac{9}{2} \cdot \frac{3}{2} = \frac{27}{4}$. This leads to the equation $x^2 - 6x + \frac{27}{4} = 0$, and multiplying by 4 produces the final form $4x^2 - 24x + 27 = 0$.

The value of $|\operatorname{adj}(\operatorname{adj}(2A))|$ is determined using the property $|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(n-1)^2}$, where $n=3$ is the order of the matrix. We first calculate the determinant of $A$ by factoring $5!$, $6!$, and $7!$ out of the rows, which simplifies the expression to

. Applying row subtractions $R_2 - R_1$ and $R_3 - R_2$ reduces the determinant to

, which evaluates to $|A| = 2$. Since $|2A| = 2^3|A| = 8(2) = 16$, the expression becomes $|2A|^4 = 16^4 = (2^4)^4 = 2^{16}$.

The determinant of the coefficient matrix, $\Delta = 18(3-a)$, identifies that the system possesses a unique solution whenever $a \neq 3$. When $a=3$, the coefficient determinant vanishes, requiring an evaluation of the numerator determinant $\Delta_x = 64 + 19b - 72a$ to determine system consistency. Substituting $a=3$ and $b=8$ results in both $\Delta=0$ and $\Delta_x=0$, which confirms the existence of infinitely many solutions. In contrast, using $a=3$ and $b=6$ yields $\Delta=0$ but produces a non-zero value for $\Delta_x$, indicating that the system is inconsistent rather than having infinitely many solutions.

The determinant of the system's coefficient matrix,

, simplifies to the expression $(\alpha-3)(\alpha+1)$. This calculation demonstrates that the system maintains a unique solution whenever $\alpha$ is neither $3$ nor $-1$. In instances where the determinant is zero, specifically at $\alpha=3$ or $\alpha=-1$, the system is either inconsistent, having no solution, or dependent, having infinitely many solutions, which is determined by the values of the determinants $D_x, D_y,$ and $D_z$.

For the specific condition $\alpha = -1$, Cramer's rule dictates that $D_x$ vanishes only when $\beta = 2$. At this precise point, all determinants including $D_x, D_y,$ and $D_z$ equal zero, resulting in a system with infinitely many solutions. Because the system possesses solutions when $\beta = 2$ and $\alpha = -1$, the assertion that there are no solutions for all $\beta \in \mathbb{R}$ when $\alpha = -1$ is mathematically incorrect. While the system is indeed inconsistent for $\alpha = -1$ when $\beta \neq 2$, and for $\alpha = 3$ when $\beta \neq 2$, the existence of an infinite solution case at $\alpha = -1$ and $\beta = 2$ identifies the second option as the invalid statement.

The fundamental property of this matrix, $P^2 = I - P$, provides a recursive way to express any higher power of $P$ as a linear combination of the identity matrix $I$ and the matrix $P$ itself. By repeatedly multiplying by $P^2$, we can determine the coefficients for successive powers of $P$. Starting from the base relation, we find $P^4 = (I - P)^2 = I - 2P + P^2$, which simplifies to $2I - 3P$ once we substitute $P^2$ with $I - P$. Continuing this process, we calculate $P^6 = P^4 P^2 = (2I - 3P)(I - P) = 2I - 2P - 3P + 3P^2$, resulting in $5I - 8P$ after simplification. Similarly, squaring $P^4$ gives $P^8 = (2I - 3P)^2 = 4I - 12P + 9P^2$, which simplifies to $13I - 21P$.

With these expressions in hand, we observe that adding $P^8$ and $P^6$ yields $(13I - 21P) + (5I - 8P) = 18I - 29P$. This matches the structure of the first given equation $P^{\alpha} + P^{\beta} = \gamma I - 29 P$, leading us to identify $\alpha = 8$, $\beta = 6$, and $\gamma = 18$. Comparing the difference $P^8 - P^6 = (13I - 21P) - (5I - 8P) = 8I - 13P$ with the second given equation $P^{\alpha} - P^{\beta} = \delta I - 13 P$ reveals that $\delta = 8$. Finally, calculating the sum $\alpha + \beta + \gamma - \delta$ gives $8 + 6 + 18 - 8$, which equals 24.

The determinant of the adjoint of a square matrix of order $m$ follows the property $|\operatorname{adj}(X)| = |X|^{m-1}$, which generalizes to $|\operatorname{adj}(\operatorname{adj}(\dots(\operatorname{adj} X)\dots))| = |X|^{(m-1)^k}$ for $k$ successive applications of the adjoint operator. Given that $A$ is a $3 \times 3$ matrix and we have three adjoints applied to the matrix $2A$, the expression $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2A))|$ evaluates to $|2A|^{(3-1)^3}$, which simplifies to $|2A|^8$. Calculating the determinant of $A$ yields $|A| = 2(4-1) - 1(2-0) = 4$, so the determinant of $2A$ is $|2A| = 2^3 |A| = 8 \times 4 = 32$. Consequently, the entire expression becomes $32^8 = (2^5)^8 = 2^{40}$. Since $16^n = (2^4)^n = 2^{4n}$, we solve for $n$ using the equality:

$2^{4n} = 2^{40} \implies 4n = 40 \implies n = 10$

The determinant of the double adjoint of a square matrix $M$ of order $m$ is defined by the relationship $\det(\operatorname{adj}(\operatorname{adj}(M))) = \det(M)^{(m-1)^2}$. Identifying the specific parameters requires solving the system of linear equations $4m + n = 22$ and $17m + 4n = 93$, which results in $m = 5$ and $n = 2$. Since the matrix $A$ is of order $m = 5$, its determinant is calculated as $|A| = m - n = 3$.

To evaluate the expression $\det( 2 \operatorname{adj}( \operatorname{adj}( 5A ) ) )$, one applies the property that for a matrix $M$ of order 5, the determinant of a scalar product is $\det(kM) = k^5 \det(M)$. This allows the expression to be written as $2^5 \det(\operatorname{adj}(\operatorname{adj}(5A)))$, which simplifies to $2^5 \cdot |5A|^{(5-1)^2} = 2^5 \cdot |5A|^{16}$. Substituting $|5A| = 5^5 |A| = 5^5 \cdot 3$ into this expression leads to $2^5 \cdot (5^5 \cdot 3)^{16} = 2^5 \cdot 5^{80} \cdot 3^{16}$. By grouping the terms $3^{16} \cdot 2^5$ as $3^{11} \cdot 3^5 \cdot 2^5$, the expression becomes $3^{11} \cdot 5^{80} \cdot 6^5$. Comparing this to $3^a 5^b 6^c$ identifies the exponents as $a = 11$, $b = 80$, and $c = 5$, yielding a sum $a + b + c = 96$.

The problem hinges on the property that for any $2 \times 2$ matrix $A$ of the form $kI$, the determinant is $\det(A^2 - A) = k^2(k - 1)^2$. Taking the transpose of the given relation $A^T = \alpha A + I$ yields $A = \alpha A^T + I$, and substituting the original expression for $A^T$ into this equation leads to $A = \alpha(\alpha A + I) + I$, which simplifies to $A(1 - \alpha^2) = (\alpha + 1)I$. Dividing by $1 + \alpha$ (since $\alpha \neq -1$) confirms that $A = \frac{1}{1 - \alpha} I$, identifying $k$ as $\frac{1}{1 - \alpha}$.

With $k = \frac{1}{1 - \alpha}$, the determinant condition $\det(A^2 - A) = 4$ becomes $(k^2 - k)^2 = 4$, which implies $k(k - 1) = \pm 2$. Substituting $\frac{1}{1 - \alpha}$ for $k$, the expression $k(k - 1)$ simplifies to $\frac{1}{1 - \alpha} \left( \frac{1}{1 - \alpha} - 1 \right) = \frac{\alpha}{(1 - \alpha)^2}$.

This leads to two cases: $\frac{\alpha}{(1 - \alpha)^2} = 2$ and $\frac{\alpha}{(1 - \alpha)^2} = -2$. The first equation rearranges to the quadratic $2\alpha^2 - 5\alpha + 2 = 0$, which has a positive discriminant and yields real roots. The second equation $2\alpha^2 - 3\alpha + 2 = 0$ has a negative discriminant, providing no real solutions. Consequently, the sum of all possible real values of $\alpha$ is determined by the sum of the roots of $2\alpha^2 - 5\alpha + 2 = 0$, which is equal to $\frac{5}{2}$.

The solvability of a system of linear equations is governed by the determinant of the coefficient matrix, which simplifies to $\Delta = (\alpha-1)^2(\alpha+2)$. Because this determinant is non-zero when $\alpha=2$, the system possesses a unique solution for any value of $\beta$, which invalidates the claim that infinitely many solutions exist for $\alpha=2$ and $\beta=-1$. In cases where $\alpha=1$ and $\beta=1$, the three equations become identical, confirming the existence of infinitely many solutions, whereas setting $\alpha=-2$ and $\beta=1$ creates a contradiction that results in no solution. Finally, by adding all three equations together when $\alpha=2$ and $\beta=1$, we obtain $4(x+y+z)=3$, which confirms that $x+y+z=\frac{3}{4}$.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, providing a method to represent the inverse of a matrix as a linear combination of the matrix itself and the identity matrix. For the matrix

, the characteristic equation is $\det(A - xI) = 0$, which evaluates to $(1-x)(10-x) - 5\lambda = 0$, or $x^2 - 11x + 10 - 5\lambda = 0$. Substituting the matrix $A$ into this equation yields $A^2 - 11A + (10 - 5\lambda)I = 0$. Multiplying this expression by $A^{-1}$ results in $A - 11I + (10 - 5\lambda)A^{-1} = 0$, which rearranges into the form $A^{-1} = \frac{-1}{10 - 5\lambda}A + \frac{11}{10 - 5\lambda}I$.

Equating this expression with $A^{-1} = \alpha A + \beta I$ gives the constants $\alpha = \frac{-1}{10 - 5\lambda}$ and $\beta = \frac{11}{10 - 5\lambda}$. The sum $\alpha + \beta$ simplifies to $\frac{10}{10 - 5\lambda}$, which is equal to $-2$. Setting $\frac{10}{10 - 5\lambda} = -2$ leads to $10 = -20 + 10\lambda$, or $\lambda = 3$. Substituting this value for $\lambda$ produces $\alpha = \frac{1}{5}$ and $\beta = -\frac{11}{5}$. Calculating the final expression $4\alpha^2 + \beta^2 + \lambda^2$ involves evaluating $4(\frac{1}{25}) + \frac{121}{25} + 9$, which simplifies to $5 + 9$, totaling 14.

A system of linear equations possesses infinitely many solutions when the determinant of its coefficient matrix is zero and the equations are consistent. Calculating the determinant along the first column gives $1(8+\alpha) - 2(-4+1) + 3(-\alpha-2) = 0$, which simplifies to $8 + \alpha + 6 - 3\alpha - 6 = 0$, resulting in $\alpha = 4$.

With $\alpha = 4$, the system comprises $x - y + z = 5$, $2x + 2y + 4z = 8$, and $3x - y + 4z = \beta$. Simplifying the second equation yields $x + y + 2z = 4$, and adding this to the first equation gives $2x + 3z = 9$. Subtracting the first equation from the third results in $2x + 3z = \beta - 5$, which must equal $9$ for the system to remain consistent and have infinitely many solutions, implying $\beta = 14$. The quadratic equation with roots $4$ and $14$ is expressed as $(x-4)(x-14) = x^2 - 18x + 56 = 0$.

For a $3 \times 3$ matrix, the determinant satisfies $|kM| = k^3|M|$ and the determinant of an adjugate matrix is $|\operatorname{adj}(M)| = |M|^2$. Starting with the expression $|\det(A) \operatorname{adj}(5 \operatorname{adj}(A^3))|$ and substituting $\det(A)=2$, we factor out the scalar $2$ to get $2^3 |\operatorname{adj}(5 \operatorname{adj}(A^3))|$. Applying the adjugate determinant property, this becomes $8 \cdot |5 \operatorname{adj}(A^3)|^2$, which expands using the scalar property to $8 \cdot (5^3 |\operatorname{adj}(A^3)|)^2 = 8 \cdot (125 \cdot |\operatorname{adj}(A^3)|)^2$. Given $|\operatorname{adj}(A^3)| = |A^3|^2 = |A|^6$, we substitute $|A|=2$ to find the inner term is $125 \cdot 2^6 = 125 \cdot 64 = 8000$. Computing the final result, $8 \cdot (8000)^2 = 8 \cdot 64 \times 10^6$ gives $512 \times 10^6$.

The provided expression for $B$ constitutes the binomial expansion of the matrix $(I - \operatorname{adj}A)^5$. Identifying the adjoint of

involves transposing the cofactor matrix, resulting in

. Subtracting this from the identity matrix $I$ gives

.

Calculating the powers of this matrix reveals

, which further leads to

. Multiplying this result by $P$ provides the final matrix:
$B = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}$
Summing all the individual entries within this matrix results in $-1 - 5 + 0 - 1$, yielding a final value of $-7$.

The matrix $A$ is constructed using the definition $a_{ij} = 2^{j-i}$, resulting in
$A = \begin{bmatrix} 1 & 2 & 4 \\ \frac{1}{2} & 1 & 2 \\ \frac{1}{4} & \frac{1}{2} & 1 \end{bmatrix}$
Squaring this matrix produces $A^2 = 3A$, which establishes a consistent recursive relationship where any higher power $A^n$ can be expressed as $3^{n-1}A$.

The sum $A^2 + A^3 + \dots + A^{10}$ can be rewritten by substituting these scalar multiples of $A$, resulting in $(3^1 + 3^2 + \dots + 3^9)A$. This expression contains a geometric series with nine terms, where the first term is $3$ and the common ratio is $3$. Applying the standard sum formula $\frac{a(r^n - 1)}{r-1}$ leads to $\frac{3(3^9 - 1)}{3 - 1}$, which simplifies to $\frac{3^{10} - 3}{2}$. Multiplying this scalar value by the matrix $A$ yields $\left( \frac{3^{10} - 3}{2} \right) A$.

A system of linear equations possesses infinitely many solutions when the coefficient determinant is zero, signifying linear dependence between the equations. Evaluating the determinant

gives $8(3) - 1(0 - \lambda) + 4(-3 - \lambda) = 12 - 3\lambda$, which indicates that $\lambda = 4$ is required for this condition to hold. Consistency for the system further requires the augmented determinant

to be zero, which results in $-2(3) - 1(-\mu) + 4(-\mu) = -6 - 3\mu = 0$, establishing that $\mu = -2$.

With the point coordinates identified as $\left(4, -2, -1/2\right)$, the perpendicular distance to the plane $8x + y + 4z + 2 = 0$ is calculated using the formula $\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting these values yields $\frac{|8(4) + 1(-2) + 4(-1/2) + 2|}{\sqrt{8^2 + 1^2 + 4^2}}$, which simplifies to $\frac{|32 - 2 - 2 + 2|}{\sqrt{81}}$. This resolves to $\frac{30}{9}$, resulting in a final distance of $\frac{10}{3}$ units.

The expansion of the expression $(A+B)^2$ yields $A^{2+}B^{2+}AB+BA$ because matrix multiplication is non-commutative. For the first condition, subtracting $A^2$ leaves $B^{2+}AB+BA$, which must equal the given constant matrix. Computing the product matrices $B^2$, $AB$, and $BA$ and summing them results in a matrix equation where each element is defined by $\alpha$ and $\beta$. Equating these components to the target matrix components leads to the solution $\beta=0$ and $\alpha_1=1$.

The second condition requires the expression $(A+B)^{2-}B^2$ to equal zero, which simplifies to the requirement that $A^{2+}AB+BA$ must be the null matrix. Calculating $A^2$ and adding it to the previously determined sum of $AB$ and $BA$ provides a new system of equations. Solving this system where all entries must be zero confirms that $\beta=0$ and yields $\alpha_2=-1$. Calculating the absolute difference between these two values, $|\alpha_1 - \alpha_2|$, results in $|1 - (-1)| = 2$.

A system of linear equations is inconsistent if the determinant of the coefficient matrix equals zero, provided that at least one of the associated determinants formed by replacing a column with the constant vector is non-zero. For the given system, the determinant of the coefficient matrix is calculated as

, which simplifies to $1(6+1) - 2(2\alpha - \alpha) + 1(\alpha - (-3\alpha)) = 7 + 2\alpha$. Setting this expression equal to zero directly forces $\alpha = -7/2$. Substituting this specific value into the determinant

results in the value $14 + 2\alpha$, which yields $14 - 7 = 7$. Because the coefficient determinant vanishes while the modified determinant remains non-zero, the system is inconsistent at $\alpha = -7/2$.

The determinant of a product of matrices is equal to the product of their determinants, while a key property of any square matrix $A$ is that the product $A \operatorname{Adj}(A)$ is equal to $\det(A)I$, where $I$ is the identity matrix. To determine the sum of the diagonal elements, which is the trace of $A$, we first expand the expression $(A+I)(\operatorname{Adj}(A)+I)$. This product simplifies to $A \operatorname{Adj}(A) + A + \operatorname{Adj}(A) + I$. Substituting $\det(A)I$ for $A \operatorname{Adj}(A)$, and using the given fact that $\det(A) = -1$, the expression becomes $-I + A + \operatorname{Adj}(A) + I$, which reduces to $A + \operatorname{Adj}(A)$.