Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The relationship between the determinant of an adjoint matrix and the original matrix, defined by $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$, offers the most efficient approach to solving this problem.

First, identify the adjoint matrix by computing the transpose of the cofactor matrix of $A$, resulting in
$\operatorname{adj}(A) = \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha-2 \end{bmatrix}$
Summing this with the first given matrix results in

.

Calculating the determinant of $B$ by expanding along the second column yields
$-3((-10)(\alpha-2) - 32) = 66$
Expanding the inner expression produces $-3(-10\alpha + 20 - 32) = 66$, which simplifies to $30\alpha + 36 = 66$, confirming that $\alpha = 1$. With the value of $\alpha$ determined, the determinant of the original matrix $A$ becomes $15(1) + 6 = 21$. Finally, raising this value to the power of $3-1$ gives $21^2 = 441$.

Any square matrix multiplied by its inverse yields the identity matrix, a principle that serves as the basis for determining the constants $\alpha$, $\beta$, and $\gamma$. First, observe that $X$ is a three-by-three nilpotent matrix with ones on the superdiagonal, which implies that $X^2$ has a single one at the top-right corner. Consequently, the expression $Y = \alpha I + \beta X + \gamma X^2$ can be explicitly written as the upper triangular matrix

.

Performing matrix multiplication between $Y$ and the provided inverse

must result in the identity matrix

. By comparing the diagonal elements, one obtains the relation $\frac{\alpha}{5} = 1$, which identifies $\alpha = 5$. Equating the entry in the first row and second column gives $\frac{\beta}{5} - \frac{2\alpha}{5} = 0$, confirming $\beta = 10$. Evaluating the top-right corner entry results in the equation $\frac{\alpha}{5} - \frac{2\beta}{5} + \frac{\gamma}{5} = 0$, which simplifies to $1 - 4 + \frac{\gamma}{5} = 0$ after substituting the known values for $\alpha$ and $\beta$, leading to $\gamma = 15$. The final value is determined by computing $(\alpha - \beta + \gamma)^2 = (5 - 10 + 15)^2$, which equals 100.

Multiplying a matrix by the standard basis vectors effectively extracts its columns, as these vectors act as identity operators that isolate specific columns from the matrix. Because the problem provides the transformations for

,

, and

, the matrix $M$ is formed by arranging these result vectors into columns to obtain

.

Expanding the matrix equation

leads to the linear system $x - z = 1$, $2x + y + z = 7$, and $3x + 2y + z = 11$. From the first equation, express $z$ as $x - 1$. Substituting this into the second equation creates $2x + y + (x - 1) = 7$, which simplifies to $y = 8 - 3x$.

Now, substitute both $z = x - 1$ and $y = 8 - 3x$ into the final equation $3x + 2y + z = 11$, resulting in $3x + 2(8 - 3x) + (x - 1) = 11$. Solving this reduces the expression to $-2x + 15 = 11$, which identifies $x = 2$. With $x$ known, calculate the remaining variables as $y = 8 - 3(2) = 2$ and $z = 2 - 1 = 1$. The sum $x + y + z$ is therefore $2 + 2 + 1 = 5$.

Recognizing that $\alpha$ acts as the complex cube root of unity, $\omega$, allows us to simplify the entry $\alpha^4$ to $\omega$. Computing the product $A \cdot A$ reveals that $A^2$ behaves as a simple permutation matrix

. Squaring this result further produces the identity matrix $I_3$, implying that the powers of $A$ repeat every four cycles such that $A^4 = I_3$.

Because the matrix follows this cyclic behavior, any high power of $A$ can be reduced using the remainder when the exponent is divided by 4. Dividing 31 by 4 leaves a remainder of 3, allowing us to express $A^{31}$ as $A^{28} \cdot A^3$. Replacing $A^{28}$ with $(A^4)^7$, which is equivalent to $I_3^7$, reduces the entire expression to $A^3$.

Representing the matrix $A$ as the sum of the identity matrix $I$ and a nilpotent matrix

allows for an efficient computation of $A^n$. Because

and $B^3$ is the zero matrix, the binomial expansion $A^n = (I+B)^n = I + nB + \frac{n(n-1)}{2}B^2$ holds perfectly. Evaluating this matrix sum produces the expression
$A^n = \begin{bmatrix} 1 & na & na + \frac{n(n-1)}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{bmatrix}$
By comparing this form with the given matrix, we establish the equations $na = 48$, $nb = 96$, and $na + \frac{n(n-1)}{2}ab = 2160$. Solving this system, we find the values $n=12$, $a=4$, and $b=8$. Consequently, the sum $n+a+b=24$.

The condition that the determinant

simplifies to the quadratic equation $\alpha^2(c-b) - \alpha(c-a) + (b-a) = 0$, which holds for $\alpha=1$ since the sum of the coefficients $(a-c) + (b-a) + (c-b)$ equals zero. Simplifying the expression $\frac{(a-c)^{2}}{(b-a)(c-b)}+\;\frac{(b-a)^{2}}{(a-c)(c-b)}+\;\frac{(c-b)^{2}}{(a-c)(b-a)}$ involves taking the common denominator $(a-c)(b-a)(c-b)$ to obtain $\frac{(a-c)^3 + (b-a)^3 + (c-b)^3}{(a-c)(b-a)(c-b)}$. Letting $X = a-c$, $Y = b-a$, and $Z = c-b$ demonstrates that $X+Y+Z=0$, and applying the algebraic identity $X^{3+}Y^{3+}Z^3 = 3XYZ$ proves the numerator simplifies to $3(a-c)(b-a)(c-b)$. Dividing this by the common denominator yields a final value of 3.

Multiplying $A$ by $B$ generates the matrix product
$AB = \begin{bmatrix} 4\alpha+4 & -4\alpha-4 \\ 2\alpha-6 & \alpha^{2+}9 \end{bmatrix}$
which must have a determinant of zero. Setting the determinant of this matrix to zero leads to the equation $(4\alpha+4)(\alpha^{2+}9) - (-4\alpha-4)(2\alpha-6) = 0$, which simplifies through factorization to $4(\alpha+1)(\alpha^{2+}2\alpha+3) = 0$. This implies either $\alpha = -1$ or $\alpha^{2+}2\alpha+3 = 0$. Since the sum of the roots for the quadratic portion is $-2$, the combined sum of all possible values for $\alpha$ is $-1 + (-2) = -3$. Finally, the absolute value of this sum is $3$.

Fundamental properties of matrix algebra dictate that for any $n \times n$ matrix $M$, the relationship $\operatorname{adj}(\operatorname{adj} M) = |M|^{n-2} M$ holds true, while the inverse of an adjoint is defined by $(\operatorname{adj} A)^{-1} = \frac{A}{|A|}$. These principles allow for the transformation of the complex nested expression into a scalar multiple of the original matrix $A$.

Determinant evaluation of $A$ yields the following:
$|A| = -4$
Substituting this into the expression for the inverse adjoint produces:
$(\operatorname{adj} A)^{-1} = \frac{A}{-4}$
Scaling this by $2$ defines the matrix $B$:
$B = 2 \left( -\frac{A}{4} \right) = -\frac{A}{2}$
The determinant of $B$ for a $3 \times 3$ matrix is calculated as:
$|B| = \left( -\frac{1}{2} \right)^3 |A| = -\frac{1}{8}(-4) = \frac{1}{2}$
Combining these values into the adjoint of the adjoint identity yields:
$\operatorname{adj}(\operatorname{adj} B) = |B| B = \frac{1}{2} \left( -\frac{A}{2} \right) = -\frac{A}{4}$
Summing the elements of the original matrix $A$ gives $1 + 1 + 2 - 2 + 0 + 1 + 1 + 3 + 5 = 12$, and applying the final scalar factor of $-\frac{1}{4}$ to this sum results in $-3$.

Solving for the matrix $B$ relies on the property that matrix multiplication is equivalent to multiplying $A$ by each column of $B$ individually. Since $B = [B_1, B_2, B_3]$, each column $B_i$ acts as a solution to a system of linear equations $AX_i = Y_i$, where $Y_i$ is the corresponding column vector of the given products.

For the first column, solving the system $A B_1 = [1, 0, 0]^T$ yields the vector $[1, -1, -1]^T$. Repeating this process for the second column, $A B_2 = [2, 3, 0]^T$ results in the vector $[2, 1, -2]^T$. The third column, obtained from $A B_3 = [3, 2, 1]^T$, produces the vector $[2, 0, -1]^T$. Assembling these column vectors gives the matrix:
$B = \begin{pmatrix} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{pmatrix}$

Calculating the determinant of $B$ yields $\alpha = 1(-1 - 0) - 2(1 - 0) + 2(2 - (-1)) = -1 - 2 + 6 = 3$. The value $\beta$ is the trace of matrix $B$, which is the sum of its diagonal elements $1 + 1 - 1 = 1$. Consequently, the final expression $\alpha^3 + \beta^3$ becomes $3^3 + 1^3 = 27 + 1 = 28$.

Expanding the determinant

along the first row gives $a(a^2 + ax) + 1(a^2x + ax^2)$, which simplifies to $a^3 + 2a^2x + ax^2$, or $a(x+a)^2$. Differentiating this expression with respect to $x$ provides the derivative $f'(x) = 2a(x+a)$.

Applying the condition $2f'(10)-f'(5)+100=0$ leads to the substitution $2(2a(10+a)) - 2a(5+a) + 100 = 0$. This equation simplifies to $4a(10+a) - 2a(5+a) + 100 = 0$, which further expands to $40a + 4a^2 - 10a - 2a^2 + 100 = 0$. Combining like terms results in the quadratic $2a^2 + 30a + 100 = 0$. Dividing the equation by 2 yields $a^2 + 15a + 50 = 0$, which has roots $a = -5$ and $a = -10$. Calculating the sum of the squares of these values gives $(-5)^2 + (-10)^2 = 125$.

The condition $A^3 = A$ implies that the entries of the matrix resulting from successive multiplication must equal the entries of the original matrix $A$. Calculating the square of the given matrix yields

. Multiplying this matrix by $A$ to obtain $A^3$ allows us to extract a system of equations by equating corresponding components, such as the entries in the first row. Setting the first entry of $A^3$ equal to the first entry of $A$ gives $2ac + 3 = 0$, and equating the second entry gives $a + 2 + 3c = 1$, which simplifies to $a + 1 + 3c = 0$. Expressing $c$ as $c = -\frac{3}{2a}$ from the first equation and substituting it into the simplified second equation results in $a + 1 + 3\left(-\frac{3}{2a}\right) = 0$, which further simplifies to $a + 1 - \frac{9}{2a} = 0$. Multiplying by $2a$ produces the quadratic equation $2a^2 + 2a - 9 = 0$. Solving for $a$ using the quadratic formula, the positive root is $a = \frac{-2 + \sqrt{4 + 72}}{4} = \frac{-1 + \sqrt{19}}{2}$. Given that $\sqrt{19}$ is approximately $4.36$, the value of $a$ is roughly $1.68$, which falls into the interval $(1, 2]$. Since $a$ belongs to $(n-1, n]$, we identify $n = 2$.

Since the summation operator is linear with respect to the determinant, the sum $\sum_{n=1}^{k} f(n)$ results in a determinant where only the first column is transformed into the sum of its original entries from $n=1$ to $k$. Using the standard power sum formulas, the first column entries become $\frac{k(k+1)}{2}$, $-\frac{k(k+1)(2k+1)}{3}$, and $-\frac{3k^2(k+1)^2}{4}$.

Evaluating this determinant simplifies the entire expression to the equation $\frac{k(k+1)(2k+1)}{2} \cdot \frac{7}{3} = 98$. This reduces further to the cubic equation $k(k+1)(2k+1) = 84$, and by testing positive integer values for $k$, it becomes clear that $k=3$ is the solution because $3 \times 4 \times 7 = 84$.

A square matrix is invertible if and only if its determinant is non-zero. To determine the conditions for invertibility, we calculate the determinant of the given matrix and ensure the result does not equal zero. By factoring out $e^t$ from the first column and $e^{-t}$ from both the second and third columns, the determinant simplifies to the product of $e^{-t}$ and the determinant of a reduced matrix.

Performing elementary row operations, specifically $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$, reduces the determinant expression to $6e^{-t}$. Since the exponential term $e^{-t}$ is strictly positive for any real number $t$, the determinant $6e^{-t}$ is always non-zero. Therefore, the matrix is invertible for all values of $t$ in the set $\mathbb{R}$.

A system of homogeneous linear equations yields non-trivial solutions if and only if the determinant of the coefficient matrix is equal to zero. By organizing the coefficients of the variables into a matrix, we obtain the requirement
$\begin{vmatrix} 1 & 1 & \sqrt{3} \\ -1 & \tan\theta & \sqrt{7} \\ 1 & 1 & \tan\theta \end{vmatrix} = 0$
Expanding this determinant along the first row involves calculating $1(\tan^2\theta - \sqrt{7}) - 1(-\tan\theta - \sqrt{7}) + \sqrt{3}(-1 - \tan\theta) = 0$. Simplifying these terms cancels out $\sqrt{7}$ and results in the quadratic form $\tan^2\theta + \tan\theta(1 - \sqrt{3}) - \sqrt{3} = 0$, which factors into the product $(\tan\theta - \sqrt{3})(\tan\theta + 1) = 0$. This confirms that non-trivial solutions occur when $\tan\theta = \sqrt{3}$ or $\tan\theta = -1$.

Within the specified interval of $[-\pi, \pi]$, these trigonometric values correspond to the angles $\theta = \frac{\pi}{3}$, $\theta = -\frac{2\pi}{3}$, $\theta = -\frac{\pi}{4}$, and $\theta = \frac{3\pi}{4}$. Summing these values, we find the total $\Sigma \theta = \frac{\pi}{3} - \frac{2\pi}{3} - \frac{\pi}{4} + \frac{3\pi}{4}$, which reduces to $\frac{\pi}{6}$. Multiplying this sum by the factor $\frac{120}{\pi}$ gives $\frac{120}{\pi} \times \frac{\pi}{6}$, resulting in 20.

Recognizing that matrix $A$ is orthogonal, meaning $AA^T = I$, allows for a significant simplification when calculating powers of $M = A^T B A$. By observing that $M^2 = (A^T B A)(A^T B A) = A^T B^2 A$, it follows by induction that $M^{2023} = A^T B^{2023} A$. This property is powerful because when we substitute this back into the expression $A M^{2023} A^T$, the terms $AA^T$ collapse into identity matrices, leaving us with $I B^{2023} I$, which reduces to simply $B^{2023}$.

The matrix

follows a predictable pattern where its $n$-th power is

. Applying this to the 2023rd power, we find

. To determine the final answer, we compute the inverse of this resulting matrix, which for an upper triangular matrix with ones on the diagonal simply requires changing the sign of the off-diagonal element, yielding

.