Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

For a system of linear equations to possess infinitely many solutions, the determinant of the coefficient matrix must be zero, ensuring the planes are linearly dependent. Calculating this determinant as

involves expanding along the first row to get $1(3a - 24) - 5(2a - 4) + 6(12 - 3) = 0$. Simplifying this expression results in $3a - 24 - 10a + 20 + 54 = 0$, which yields $-7a + 50 = 0$, determining that $a = \frac{50}{7}$.

Because the system is consistent and has infinitely many solutions, the determinant formed by replacing the third column with the constants must also be zero. Expanding

gives $1(3b - 42) - 5(2b - 7) + 4(12 - 3) = 0$, which simplifies to $3b - 42 - 10b + 35 + 36 = 0$. This reduces to $-7b + 29 = 0$, identifying that $b = \frac{29}{7}$. Comparing these results, we find that $a - b = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$, meaning the point $(a, b)$ satisfies the linear relationship $x - y = 3$.

The indefinite integral $\int \frac{7 x^{10}+9 x^8}{(1+x^{2+}2 x^9)^2} \, dx$ simplifies by dividing the numerator and denominator by $x^{18}$ to $\int \frac{7 x^{-8}+9 x^{-10}}{(x^{-9}+x^{-7}+2)^2} \, dx$, which is the derivative of $-(x^{-9}+x^{-7}+2)^{-1}$ with respect to $x$. Performing the integration, we obtain $f(x) = \frac{1}{x^{-9}+x^{-7}+2} + C = \frac{x^9}{1+x^{2+}2x^9} + C$. Given $\lim_{x\to 0}f(x)=0$, the constant of integration $C$ must be 0, yielding $f(x) = \frac{x^9}{1+x^{2+}2x^9}$.

Evaluating the derivative at $x=1$ using the quotient rule, we find $f'(1) = \frac{(1+1^{2+}2(1)^9)(9(1)^8) - (1)^9(2(1)+18(1)^8)}{(1+1^{2+}2(1)^9)^2} = \frac{4(9) - 20}{16} = 1$. The matrix $A$ is therefore

, and its determinant is $|A| = 1\left(\frac{1}{4}(4) - \alpha^2(1)\right) = 1 - \alpha^2$.

Using the identity $|\operatorname{adj}(\operatorname{adj} A)| = |A|^{(n-1)^2}$ for an $n \times n$ matrix, where $n=3$, the given condition $|B|=81$ becomes $|A|^4 = 81$. This implies $|A| = \pm 3$, resulting in the equations $1 - \alpha^2 = 3$ or $1 - \alpha^2 = -3$. Solving the latter yields $\alpha^2 = 4$.

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation, defined by $\det(M - \lambda I) = 0$. For

, the characteristic equation is

. Equating coefficients with $A^2 - 4A + 2I = 0$ gives $\alpha + 2 = 4$ and $2\alpha - 2 = 2$, both implying $\alpha = 2$, so

. Similarly, for

, the characteristic equation

compared with $B^2 - 3B + I = 0$ gives $\beta = 2$, so

.

With

, calculation yields

and

. For

, we find

and

. Subtracting these matrices results in

. Utilizing the property $\det(\operatorname{adj}(M)) = \det(M)^{n-1}$ for a $2 \times 2$ matrix, where $n=2$, the expression becomes $(\det(A^3 - B^3))^{2(2-1)}$. Calculating the determinant $\det(A^3 - B^3) = (15)(7) - (20)(6) = 105 - 120 = -15$, the final value is $(-15)^2 = 225$.

A homogeneous system of linear equations possesses infinitely many solutions if and only if the determinant of its coefficient matrix equals zero. Evaluating the determinant of the coefficient matrix gives the equation
$\begin{vmatrix} 1 & 2 & t \\ 6 & 1 & 5t \\ 3 & t^2 & f(t) \end{vmatrix} = 0$
which simplifies to the function $f(t) = \frac{t^{3+}12t}{11}$. Differentiating this expression with respect to $t$ yields the derivative $f'(t) = \frac{3t^{2+}12}{11}$. Since the term $3t^2$ is always non-negative for any real number $t$, the entire derivative $f'(t)$ remains strictly positive for all values of $t$, confirming that the function is strictly increasing on $R$.

According to the Cayley-Hamilton theorem, the matrix

satisfies its characteristic equation $\lambda^2 - 7\lambda + 1 = 0$, yielding the relation $A^2 = 7A - I$. Factoring the expression $A^{2025} - 3 A^{2024} + A^{2023}$ as $A^{2023}(A^2 - 3A + I)$ allows for the substitution of $7A - I$ for $A^2$, resulting in $A^{2023}(4A) = 4A^{2024}$. Given $\det(A) = (2)(5) - (3)(3) = 1$, the determinant property $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix with $n=2$ yields $\det(4A^{2024}) = 4^2 (\det A)^{2024} = 16(1)^{2024} = 16$.

Expanding the given matrix expression using the binomial expansion, we have $(A+I)^{3+}(A-I)^{3-}6A = (A^3 + 3A^2 + 3A + I) + (A^3 - 3A^2 + 3A - I) - 6A = 2A^3$. The matrix $A$ represents a rotation matrix about the $y$-axis, which satisfies the property

. Setting $A^2 = A^T$ leads to $\cos(2\theta) = \cos\theta$ and $\sin(2\theta) = -\sin\theta$, which for $\theta \in (0, \pi)$ yields $\theta = 2\pi/3$. Substituting this value into the matrix power formula for $A^3$ results in

. The expression therefore simplifies to $2I$, and the sum of the diagonal elements is 6.

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 determinant formed by replacing a variable column with the constant terms must also vanish. Calculating the determinant of the coefficient matrix yields

, which must equal zero for the system to lack a unique solution.

The requirement for consistency involves setting the determinant where the third column is replaced by the constant vector to zero, specifically

. Solving this equation immediately determines $\alpha = -\frac{19}{9}$.

Substituting this value into the equation $3\alpha + 4\beta - \alpha\beta + 3 = 0$ results in $3(-\frac{19}{9}) + 4\beta - (-\frac{19}{9})\beta + 3 = 0$. Clearing the denominators by multiplying by 9 produces $-57 + 36\beta + 19\beta + 27 = 0$, which simplifies to $55\beta = 30$, giving $\beta = \frac{6}{11}$. Evaluating the final expression $22\beta - 9\alpha$ provides $22(\frac{6}{11}) - 9(-\frac{19}{9}) = 12 + 19 = 31$.

Matrix multiplication follows the fundamental rule where each entry in the product matrix is derived from the dot product of a row from the first matrix and a column from the second. Constructing matrix $A$ based on the given element definition $a_{ij} = (\sqrt{2})^{i+j}$ results in the following structure:
$A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix}$
To find the sum of the elements in the third row of $A^2$, calculate the dot product of the third row of $A$, which is $[4 \quad 4\sqrt{2} \quad 8]$, with each of the three columns of $A$. The resulting elements for the third row are $4(2) + 4\sqrt{2}(2\sqrt{2}) + 8(4) = 56$ for the first position, $4(2\sqrt{2}) + 4\sqrt{2}(4) + 8(4\sqrt{2}) = 56\sqrt{2}$ for the second, and $4(4) + 4\sqrt{2}(4\sqrt{2}) + 8(8) = 112$ for the third. Adding these individual components yields a total of $168 + 56\sqrt{2}$. Identifying this with the form $\alpha + \beta\sqrt{2}$ gives $\alpha = 168$ and $\beta = 56$, and their sum is $168 + 56 = 224$.

The behavior of a matrix raised to a power becomes significantly simpler when the matrix satisfies the idempotent property, $A^2 = A$. To determine the entries of $A$, the condition $\det(A) = \alpha\beta - (-6) = 0$ along with $\alpha+\beta=1$ leads to the quadratic roots $\alpha=3$ and $\beta=-2$. Direct calculation verifies that

, confirming the idempotent property. Because $A^k = A$ for all positive integers $k$, the binomial expansion of $(I+A)^8$ simplifies to $I + \binom{8}{1} A + \binom{8}{2} A^2 + \dots + \binom{8}{8} A^8$, which collapses into $I + (\binom{8}{1} + \binom{8}{2} + \dots + \binom{8}{8}) A$.

Given that the sum of the binomial coefficients $\sum_{k=1}^8 \binom{8}{k}$ equals $2^8 - 1 = 255$, the entire expression reduces to $I + 255A$. Computing this sum yields the final matrix result:
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + 255 \begin{pmatrix} 3 & -1 \\ 6 & -2 \end{pmatrix} = \begin{pmatrix} 1 + 765 & -255 \\ 1530 & 1 - 510 \end{pmatrix} = \begin{pmatrix} 766 & -255 \\ 1530 & -509 \end{pmatrix}$

A matrix $A$ satisfying the condition $X^{\top} A X = O$ for all nonzero vectors $X$ is necessarily a skew-symmetric matrix. This property implies that all diagonal elements are zero and $A^{\top} = -A$, allowing us to represent the matrix as

for unknown constants $p$, $q$, and $r$.

The matrix transformations provided establish systems of linear equations for these variables. Using

, we find $p+q=1$, $-p+r=4$, and $q+r=5$. The second transformation

provides the additional constraints $2p+q=0$ and $q+2r=8$. Solving these equations, we obtain $p=-1$, $q=2$, and $r=3$, which defines the matrix as

.

With this matrix defined, the expression $2(A+I)$ becomes

. Calculating the determinant of this resulting matrix yields $120$. Since the property of the adjoint determinant for a $3 \times 3$ matrix is $\det(\operatorname{adj}(M)) = (\det M)^2$, the determinant in this case is $120^2$. Prime factorization reveals $120^2 = (2^3 \cdot 3 \cdot 5)^2 = 2^6 \cdot 3^2 \cdot 5^2$. Comparing this to the given expression $2^{\alpha} 3^{\beta} 5^{\gamma}$ results in $\alpha = 6$, $\beta = 2$, and $\gamma = 2$. Consequently, the final sum $\alpha^2 + \beta^2 + \gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$.

The behavior of the determinant under scalar multiplication and the properties of the adjoint matrix are fundamental to solving this problem. For an $n \times n$ matrix $M$, we apply the rules $|kM| = k^n |M|$ and $|\operatorname{adj}(M)| = |M|^{n-1}$. Since the matrix $A$ is $3 \times 3$, we have $n=3$, which means $|kM| = k^3 |M|$ and $|\operatorname{adj}(M)| = |M|^2$. Applying these to the expression $|2 \operatorname{adj}(3 A \operatorname{adj}(2 A))|$, we first factor out the scalar $2$ to get $2^3 |\operatorname{adj}(3 A \operatorname{adj}(2 A))|$, which then simplifies to $2^3 |3 A \operatorname{adj}(2 A)|^2$.

To evaluate the inner term $|3 A \operatorname{adj}(2 A)|$, we identify it as the product of two matrices, $3A$ and $\operatorname{adj}(2 A)$. Applying the property $|3 A| = 3^3 |A|$, we get $3^3 |A| |\operatorname{adj}(2 A)|$. Because $|\operatorname{adj}(2 A)| = |2 A|^2$ and $|2 A| = 2^3 |A| = 8 |A|$, the term becomes $3^3 |A| (8 |A|)^2$, which simplifies to $27 |A| \cdot 64 |A|^2$, or $1728 |A|^3$. Squaring this result gives $(27 \cdot 64 |A|^3)^2 = 3^6 \cdot 2^{12} \cdot |A|^6$. Multiplying this by the outer factor $2^3$ results in $2^{15} \cdot 3^6 \cdot |A|^6$. Given $|A| = 5$, the full expression is $2^{15} \cdot 3^6 \cdot 5^6$. Summing the exponents $\alpha = 15$, $\beta = 6$, and $\gamma = 6$ results in $27$.

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the determinants formed by replacing a column with the constants vector must also vanish. This consistency requirement leads to the determinant condition:

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & m \\ 1 & 4 & m^2 \end{vmatrix} = 0$

Expanding this expression yields $1(2m^2 - 4m) - 1(m^2 - m) + 1(4 - 2) = 0$, which simplifies to the quadratic equation $m^2 - 3m + 2 = 0$. Solving this gives the roots $\alpha = 1$ and $\beta = 2$.

With these values, the summation $\sum_{n=1}^{10} (n^1 + n^2)$ splits into the sum of the first ten integers and the sum of their squares. Using standard formulas, the sum of integers is $\frac{10(11)}{2} = 55$, and the sum of squares is $\frac{10(11)(21)}{6} = 385$. Adding these values together produces 440.

The matrix equation $A^2(A-2I) - 4(A-I) = 0$ rearranges to $A^3 = 2A^2 + 4A - 4I$. Multiplying both sides by $A$ gives $A^4 = 2A^3 + 4A^2 - 4A$, which simplifies by substituting the expression for $A^3$:
$A^4 = 2(2A^2 + 4A - 4I) + 4A^2 - 4A = 8A^2 + 4A - 8I$

Continuing the calculation to determine $A^5$ involves multiplying this expression by $A$ to get $8A^3 + 4A^2 - 8A$, and replacing $A^3$ once more yields:
$A^5 = 8(2A^2 + 4A - 4I) + 4A^2 - 8A = 20A^2 + 24A - 32I$

Comparing this with $A^5 = \alpha A^2 + \beta A + \gamma I$ establishes $\alpha = 20$, $\beta = 24$, and $\gamma = -32$. The final sum of these constants, $20 + 24 - 32$, equals 12.

The characteristic equation of the matrix

is $\det(A-\lambda I) = \lambda^2 - 2\lambda + 1 = 0$, indicating a repeated eigenvalue of $1$. By the Cayley-Hamilton theorem, $A$ satisfies $(A-I)^2 = 0$. By letting

, it follows that $N^2 = 0$, which simplifies matrix powers significantly since $A^k = (I+N)^k = I + kN$ for any integer $k$.

Applying this simplification to the equation $A^{m^2} + A^m = 3I - A^{-6}$, we substitute $I + m^2N$ for $A^{m^2}$, $I + mN$ for $A^m$, and $I - 6N$ for $A^{-6}$. This transforms the equation into $(I + m^2N) + (I + mN) = 3I - (I - 6N)$, which simplifies to $2I + (m^{2+}m)N = 2I + 6N$. Comparing the coefficients of the nilpotent matrix $N$ yields the quadratic equation $m^2 + m = 6$, or $m^2 + m - 6 = 0$. Factoring this expression gives $(m+3)(m-2) = 0$, which identifies two distinct integer solutions for $m$.

The core principle here is that multiplying a matrix by a vector produces a linear combination of that matrix's columns, where the vector's components act as weights. By examining the first transformation
$A \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
we immediately identify the second column of $A$ as the vector

, which means the middle entry of the second row, $a_{22}$, is $0$.

We are specifically interested in the second row of the matrix, which generates the second component of the result vector. Representing the elements of the second row as $a_{21}, a_{22}, a_{23}$, the second component of any product $A\mathbf{x}$ is calculated as $a_{21}x_1 + a_{22}x_2 + a_{23}x_3$. Substituting the known value $a_{22} = 0$, this simplifies to $a_{21}x_1 + a_{23}x_3$. Applying this logic to the provided input-output pairs, the second component of
$A \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$
yields the linear equation $4a_{21} + 3a_{23} = 1$. Similarly, the second component of
$A \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
yields $2a_{21} + 2a_{23} = 0$. Solving this system, the second equation implies $a_{21} = -a_{23}$. Substituting this into the first equation, we find $4(-a_{23}) + 3a_{23} = 1$, which simplifies to $-a_{23} = 1$, confirming that $a_{23} = -1$.

Infinitely many solutions for a system of linear equations exist only when the determinant of the coefficient matrix equals zero and the system is consistent. Evaluating the determinant
$\begin{vmatrix} 1 & 5 & -1 \\ 4 & 3 & -3 \\ 24 & 1 & \lambda \end{vmatrix} = -17\lambda - 289$
leads directly to $\lambda = -17$. Ensuring the system is consistent requires finding $m$ such that the augmented determinant is zero, identifying $m=45$. Representing the system in terms of an integer parameter $y=k$ allows for expressing the other variables as $x = 4 + 12k$ and $z = 3 + 17k$.

Imposing the constraint $7 \leq x+y+z \leq 77$ requires calculating the total sum in terms of $k$. Combining the parametric forms yields $(4+12k) + k + (3+17k) = 7 + 30k$. The inequality $7 \leq 7 + 30k \leq 77$ simplifies to $0 \leq 30k \leq 70$, which constrains $k$ to $0 \leq k \leq 2.33$. Only three integer values for $k$ exist within this range, indicating three possible solutions.

The property that the adjugate of a matrix product involves the product of adjugates allows for significant simplification of $B$. Specifically, using the identity $\operatorname{adj}(A^2) = (\operatorname{adj} A)^2$, the expression $\operatorname{adj} (A \operatorname{adj} (A^2))$ simplifies because $A \operatorname{adj} (A^2)$ becomes $A(\operatorname{adj} A)^2$. Since $\operatorname{adj} A = |A|A^{-1}$, this further reduces to $A(|A|A^{-1})^2 = |A|^2 A^{-1}$. Given $|A| = -1$, the matrix $A \operatorname{adj} (A^2)$ simplifies to $A^{-1}$. The matrix $B$ is the inverse of $\operatorname{adj}(A^{-1})$, and using the property $\operatorname{adj}(M) = |M|M^{-1}$, we find $\operatorname{adj}(A^{-1}) = |A^{-1}|(A^{-1})^{-1} = \frac{1}{|A|} A$. Thus, $B = (\frac{1}{-1} A)^{-1} = -A^{-1}$.