Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations possesses a unique solution if and only if the determinant of the coefficient matrix, $\Delta$, is non-zero. For the given system, calculating this determinant yields $\Delta = 6-2\alpha$, which implies that the system is singular whenever $\alpha = 3$. Since a unique solution requires $\Delta \neq 0$, the claim that the system possesses a unique solution when $\alpha = 3$ is mathematically inconsistent.

In scenarios where $\alpha = 3$, the system is either inconsistent or dependent, with its specific behavior determined by other factors like $\Delta z = \beta - 14$. When $\alpha = 3$ and $\beta = 14$, the system admits infinitely many solutions, whereas $\alpha = 3$ combined with $\beta \neq 14$ results in no solution. Consequently, because the condition $\alpha = 3$ fundamentally prevents the existence of a unique solution, the statement proposing otherwise is invalid.

A system of linear equations possesses infinitely many solutions when the equations are linearly dependent, meaning they essentially represent consistent, redundant constraints rather than defining a unique point of intersection. We can determine the values of $\lambda$ and $\mu$ by reducing the system to eliminate variables. Starting with the first equation $x+2y+3z=3$, we multiply it by $4$ to get $4x+8y+12z=12$ and subtract the second equation $4x+3y-4z=4$, which yields the simplified relationship $5y+16z=8$. Similarly, we eliminate $x$ by multiplying the second equation by $2$ to obtain $8x+6y-8z=8$ and subtracting the third equation $8x+4y-\lambda z=9+\mu$, resulting in the expression $2y+(\lambda-8)z=-1-\mu$.

For the system to have infinitely many solutions, the two resulting linear equations in $y$ and $z$ must be proportional, representing the same line. Equating the ratios of their respective coefficients and constants gives $\frac{5}{2} = \frac{16}{\lambda-8} = \frac{8}{-1-\mu}$. Solving the first equality, $5(\lambda-8) = 32$ leads to $5\lambda-40=32$, which simplifies to $5\lambda = 72$, or $\lambda = \frac{72}{5}$. Applying the same ratio to the second part, $5(-1-\mu) = 16$ results in $-5-5\mu = 16$, which further simplifies to $-5\mu = 21$, or $\mu = \frac{-21}{5}$. The resulting ordered pair is thus $(\frac{72}{5}, \frac{-21}{5})$.

Because the summation is applied to the determinant and the variable $k$ appears exclusively in the first row, the summation operator can be distributed directly into those specific elements without affecting the entries of the other rows. Summing the terms $1$, $2k$, and $2k-1$ from $1$ to $n$ yields $n$, $n^{2+}n$, and $n^2$ respectively, which updates the determinant's first row. The resulting expression is:

$\begin{vmatrix} n & n^{2+}n & n^2 \\ n & n^{2+}n+2 & n^2 \\ n & n^{2+}n & n^{2+}n+2 \end{vmatrix} = 96$

Subtracting the first row from the second row and the first row from the third row simplifies the matrix significantly. Performing the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ reduces the determinant to an upper triangular form:

$\begin{vmatrix} n & n^{2+}n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2 \end{vmatrix} = 96$

Evaluating this determinant by expanding along the first column yields $n \cdot (2(n+2)) = 96$, which simplifies to $n(n+2) = 48$. Factoring the quadratic equation $n^2 + 2n - 48 = 0$ gives $(n+8)(n-6) = 0$, leading to the result $n = 6$.

A symmetric matrix of order 3 is characterized by the property that its transpose is equal to the matrix itself, which implies that entries reflected across the main diagonal must be identical. In a 3x3 grid, this symmetry requires the three diagonal elements and the three entries in the upper triangular portion to be chosen independently, while the remaining three entries in the lower triangular positions are automatically fixed to match their counterparts.

Since there are 6 such independent positions available, and each position can be filled using any of the 10 digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, we apply the fundamental counting principle to determine the total number of possible matrices. By multiplying the number of available choices for each of these 6 independent spots, we perform the calculation $10 \times 10 \times 10 \times 10 \times 10 \times 10$, resulting in $10^6$ distinct symmetric matrices.

A system of linear equations possesses infinitely many solutions when the equations are linearly dependent, meaning the third equation can be expressed as a linear combination of the other two. By examining the coefficients of the first two equations, $7x + 11y + \alpha z = 13$ and $5x + 4y + 7z = \beta$, we can determine the multipliers $k_1$ and $k_2$ such that $k_1(7, 11) + k_2(5, 4) = (175, 194)$. Solving this system of two variables yields $k_1 = 10$ and $k_2 = 21$. Applying these same multipliers to the $z$ coefficients and the constant terms in the third equation gives $10\alpha + 21(7) = 57$ and $10(13) + 21\beta = 361$. Solving $10\alpha + 147 = 57$ leads to $10\alpha = -90$, so $\alpha = -9$. Similarly, $130 + 21\beta = 361$ simplifies to $21\beta = 231$, resulting in $\beta = 11$. Calculating the required sum, $-9 + 11 + 2$, yields the final result of $4$.

Determining the consistency of a linear system involves checking the dependencies between equations, which is achieved by expressing two variables in terms of the third to observe the constraints on the coefficients. By subtracting the first equation from the third, we find $y = 8 - 2z$, which allows us to deduce $x = z - 2$ using the first equation. Substituting these expressions into the second equation yields the constraint $z(\alpha - 2\beta + 7) = 2\alpha - 8\beta + 3$. The system possesses a unique solution when $\alpha - 2\beta + 7 \neq 0$, whereas it has no solution if $\alpha - 2\beta + 7 = 0$ but $2\alpha - 8\beta + 3 \neq 0$, and infinitely many solutions only when both expressions simultaneously equal zero.

Testing these conditions confirms the system's behavior under various parameter values. If $\alpha = \beta = 7$, the condition $\alpha - 2\beta + 7 = 0$ holds while the constant term $2\alpha - 8\beta + 3$ evaluates to $-39$, which verifies that the system has no solution. Regarding the unique solution condition, setting $\alpha = \beta$ and $\alpha \neq 7$ ensures that $\alpha - 2\beta + 7 = 7 - \alpha \neq 0$, which is consistent. Infinitely many solutions occur only at the specific point $(\alpha, \beta) = (-12.5, -2.75)$, obtained by solving the simultaneous equations $\alpha - 2\beta + 7 = 0$ and $2\alpha - 8\beta + 3 = 0$; substituting these values into the line $x + 2y + 18 = 0$ (where $x$ and $y$ represent $\alpha$ and $\beta$) confirms the point lies on that line. Because infinitely many solutions occur only at this single coordinate, it is impossible for every point on the line $x - 2y + 7 = 0$ to satisfy the requirement for an infinite number of solutions.

A system of linear equations possesses infinitely many solutions only when the determinant of the coefficient matrix is zero, signifying that the equations are linearly dependent, and the determinants formed by substituting the constants for the variable columns also vanish to ensure consistency. Evaluating the coefficient determinant

yields the equation $1(15-4) - 1(6-2) + a(4-5) = 7-a = 0$, establishing that $a$ must be 7.

For the system to remain consistent, the determinant formed by replacing the $z$ column with the constant vector,

, must also be set to zero. Expanding this determinant gives $1(15-12) - 1(6-6) + b(4-5) = 3 - b = 0$, which confirms $b = 3$. Calculating the expression $2a + 3b$ using these values yields $2(7) + 3(3)$, resulting in 23.

Matrices that satisfy the condition $A^2=A$ are known as idempotent matrices, a property which implies that raising the matrix to any positive integer power yields the original matrix $A$. Verifying the product of $A$ with itself confirms that this specific matrix is idempotent, which greatly simplifies the binomial expansion of $(A+I)^{11}$. Applying the binomial theorem allows us to write $(A+I)^{11}$ as the summation $\sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k}$. Because $A^k$ equals $A$ for all $k \geq 1$ and the identity matrix $I$ raised to any power remains $I$, the expression simplifies to $(\sum_{k=1}^{11} \binom{11}{k}) A + \binom{11}{0} I$.

Given that the sum of binomial coefficients $\sum_{k=0}^{11} \binom{11}{k}$ equals $2^{11}$, we subtract the term for $k=0$, which is $1$, to find that $\sum_{k=1}^{11} \binom{11}{k} = 2^{11}-1 = 2047$. Substituting this into the previous expression results in $2047 A + I$. To determine the sum of the diagonal elements, we calculate the trace of this resulting matrix, which is $2047 \cdot \text{tr}(A) + \text{tr}(I)$. Since the diagonal elements of $A$ are $1$, $4$, and $-3$, their sum is $2$, and the sum of the diagonal elements of the $3 \times 3$ identity matrix $I$ is $3$. Multiplying $2047$ by $2$ and adding $3$ produces $4094 + 3$, which results in $4097$.

A system of linear equations has infinitely many solutions when its coefficient matrix determinant equals zero. Evaluating the determinant

yields $1(10) - 1(7) - k(1) = 0$, which simplifies to $3 - k = 0$ and identifies $k = 3$. Substituting $k = 3$ into the second system provides the following equations:
$4x + 5y = 7$
$7x + 8y = 10$
Subtracting the first equation from the second results in $3x + 3y = 3$, demonstrating that the system has a unique solution satisfying $x + y = 1$.

The property of a symmetric matrix requires it to take the form

, with the given determinant condition $|A| = ac - b^2 = 2$. By carrying out the matrix multiplication provided in the problem, we equate the resulting entries to obtain the linear relationships $2a + b = 1$ and $2b + c = 2$.

$\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} = \begin{bmatrix} 2a+b & 2b+c \\ 3a+\frac{3b}{2} & 3b+\frac{3c}{2} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$

Expressing $a$ as $\frac{1-b}{2}$ and $c$ as $2-2b$, we substitute these into the determinant equation to get $\left(\frac{1-b}{2}\right) (2-2b) - b^2 = 2$. This simplifies to $(1-b)^2 - b^2 = 2$, which implies $1 - 2b = 2$, yielding $b = -\frac{1}{2}$. Applying this value, we find $a = \frac{3}{4}$ and $c = 3$. The sum of the diagonal elements $s$ is calculated as $a + c = \frac{3}{4} + 3 = \frac{15}{4}$. Moving to the variables $\alpha$ and $\beta$, we substitute the known values to get $\alpha = 3(\frac{3}{4}) + \frac{3}{2}(-\frac{1}{2}) = \frac{9}{4} - \frac{3}{4} = \frac{3}{2}$ and $\beta = 3(-\frac{1}{2}) + \frac{3}{2}(3) = -\frac{3}{2} + \frac{9}{2} = 3$. Finally, computing the requested expression $\frac{\beta s}{\alpha^2}$ gives $\frac{3 \times \frac{15}{4}}{(\frac{3}{2})^2} = \frac{45/4}{9/4} = 5$.

The symmetry of matrix $P$ is revealed by taking the transpose of $P^T = aP + (a-1)I$ to get $P = aP^T + (a-1)I$. Subtracting these two expressions leads to $(1+a)(P^T - P) = 0$, and since $a > 1$, the factor $(1+a)$ is non-zero, requiring $P = P^T$. This confirms that $P$ is symmetric, allowing us to substitute $P^T$ with $P$ in the original equation to get $P = aP + (a-1)I$, which simplifies to $P = -I$. For a $3 \times 3$ matrix, the determinant $|P| = |-I| = -1$, and applying the property $|\text{Adj } P| = |P|^{n-1}$ results in $|\text{Adj } P| = (-1)^{3-1} = 1$.

The nested adjoint operation of a matrix can be simplified using the general property $|\operatorname{adj}(\operatorname{adj} M)| = |M|^{(n-1)^2}$, where $M$ is an $n \times n$ matrix. For the given $3 \times 3$ matrix $A$, substituting $n=3$ simplifies this identity to $|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^4 = |A|^8$. Determining the determinant of $A$ is therefore the primary step required to find the final value.

By writing the logarithmic entries of $A$ as fractions, one can factor out the inverse of the product $\log x \cdot \log y \cdot \log z$ from the respective rows to simplify the determinant expression to

. Canceling the logarithmic terms leaves the determinant of

. Evaluating this determinant produces $1(6 - 1) - 1(3 - 1) + 1(1 - 2) = 2$, and substituting this result into the previous relation yields $2^8$.

To determine if a matrix expression is symmetric or skew-symmetric, we check if its transpose equals itself or its negative, utilizing the rules $(AB)^T = B^T A^T$ and $(A^n)^T = (A^T)^n$. Given $A^T = A$ and the skew-symmetric properties $B^T = -B$ and $C^T = -C$, we can evaluate the two expressions. For the first expression $M = A^{13} B^{26} - B^{26} A^{13}$, the transpose is $M^T = (B^T)^{26} (A^T)^{13} - (A^T)^{13} (B^T)^{26}$. Substituting the given properties leads to $B^{26} A^{13} - A^{13} B^{26}$, which is $-M$, proving this expression is skew-symmetric. For the second expression $N = A^{26} C^{13} - C^{13} A^{26}$, the transpose is $N^T = (C^T)^{13} (A^T)^{26} - (A^T)^{26} (C^T)^{13}$. Substituting $A^T = A$ and $C^T = -C$ results in $(-C)^{13} A^{26} - A^{26} (-C)^{13}$, which simplifies to $-C^{13} A^{26} + A^{26} C^{13}$ because $(-1)^{13} = -1$. This simplifies further to $A^{26} C^{13} - C^{13} A^{26}$, which is equal to $N$, confirming that this expression is symmetric.

Matrices of this specific structure represent rotation transformations, which can be expressed in the form

with an angle $\alpha = 60^{\circ}$. A key property of these matrices is that raising them to the power of $n$ multiplies the angle by $n$, yielding

.

Applying this to the given problem, $A^{30}$ results in an angle of $30 \times 60^{\circ} = 1800^{\circ}$, which simplifies to $0^{\circ}$ modulo $360^{\circ}$, making $A^{30} = I$. Similarly, $A^{25}$ results in an angle of $25 \times 60^{\circ} = 1500^{\circ}$, which simplifies to $60^{\circ}$ modulo $360^{\circ}$, so $A^{25} = A$. Substituting these values into the expression $A^{30} + A^{25} - A$ results in $I + A - A$, which reduces to the identity matrix $I$.

The determinant of the adjoint of an $n \times n$ matrix $M$ is defined as $|\operatorname{Adj}(M)| = |M|^{n-1}$, and multiplying a matrix by a scalar $k$ affects its determinant according to the rule $|k M| = k^n |M|$. By applying these two principles systematically, the nested determinants can be evaluated from the inside out. Starting with the inner term $2A^{-1}$, and given $|A|=2$, the determinant is $|2A^{-1}| = 2^n |A^{-1}| = 2^n \cdot \frac{1}{2} = 2^{n-1}$. Applying the adjoint property, $|\operatorname{Adj}(2A^{-1})| = |2A^{-1}|^{n-1} = (2^{n-1})^{n-1} = 2^{(n-1)^2}$.

Next, consider the matrix $2 \cdot \operatorname{Adj}(2A^{-1})$. Its determinant is $2^n \cdot |\operatorname{Adj}(2A^{-1})| = 2^n \cdot 2^{(n-1)^2} = 2^{n + n^2 - 2n + 1} = 2^{n^2 - n + 1}$. Finally, taking the adjoint of this resulting matrix gives a determinant of $|2 \cdot \operatorname{Adj}(2A^{-1})|^{n-1}$, which expands to $(2^{n^2 - n + 1})^{n-1} = 2^{(n-1)(n^2 - n + 1)}$. Equating this expression to $2^{84}$ leads to the algebraic condition $(n-1)(n^2 - n + 1) = 84$. Testing integer values, when $n = 5$, we have $4 \times (25 - 5 + 1) = 4 \times 21$, which equals 84, confirming $n = 5$.

A system of linear equations fails to have a solution when the determinant of the coefficient matrix is zero and the equations are not linearly dependent. Evaluating the determinant for the provided system yields the polynomial $(\lambda + 2)(\lambda - 1)^2$, which equals zero precisely when $\lambda$ is $1$ or $-2$.

Substituting $\lambda = 1$ into the system produces three identical equations, which implies the presence of infinitely many solutions rather than inconsistency. Conversely, when $\lambda = -2$, the equations result in a contradiction, confirming that the system is indeed inconsistent. Since the set $S$ contains only $\lambda = -2$, evaluating the expression $\left|\lambda\right|^{2+}\left|\lambda\right|$ with this value leads to $4 + 2$, giving a total of $6$.

Any power of a matrix can be reduced to a linear combination of the matrix and the identity matrix by repeatedly applying the defining relation $A^2 = 3A + \alpha I$. Starting with $A^2 = 3A + \alpha I$, we multiply both sides by $A$ to obtain $A^3 = 3A^2 + \alpha A$. Substituting the original expression for $A^2$ into this equation gives $A^3 = 3(3A + \alpha I) + \alpha A$, which simplifies to $A^3 = (9 + \alpha)A + 3\alpha I$. Continuing this process to determine $A^4$, we multiply $A^3$ by $A$ to get $A^4 = (9 + \alpha)A^2 + 3\alpha A$.

Replacing $A^2$ one last time yields $A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A$, which expands to $A^4 = (27 + 3\alpha)A + (9\alpha + \alpha^2)I + 3\alpha A$, or $A^4 = (27 + 6\alpha)A + (\alpha^2 + 9\alpha)I$. By equating the coefficients of $A$ and $I$ with the provided expression $A^4 = 21A + \beta I$, we identify that $27 + 6\alpha = 21$, which leads to $6\alpha = -6$ or $\alpha = -1$. Substituting $\alpha = -1$ into the constant term $\beta = \alpha^2 + 9\alpha$ results in $\beta = (-1)^2 + 9(-1)$, giving a final value of $\beta = -8$.

For the matrix

with determinant $d = mq - np$, the adjugate is given by

. Forming the matrix $A - d(\text{Adj } A)$ yields

, and setting its determinant equal to zero results in the equation $(m-qd)(q-md) - np(1+d)^2 = 0$. Expanding this expression leads to $(mq - np) + d^2(mq - np) - d(m^2 + q^2 + 2np) = 0$, where substituting $mq - np = d$ and using the identity $m^2 + q^2 + 2np = (m+q)^2 - 2d$ simplifies the equation to $d + d^3 - d((m+q)^2 - 2d) = 0$. Dividing by the non-zero determinant $d$ results in $1 + d^2 = (m+q)^2 - 2d$, which simplifies to the final relationship $(1+d)^2 = (m+q)^2$.

Properties of determinants for a $3 \times 3$ matrix, such as $|kA| = k^3 |A|$ and $|\operatorname{adj}(M)| = |M|^2$, allow for an efficient evaluation of this expression. First, the determinant of the inner scalar term is $|3A| = 3^3 |A| = 27 \times 2 = 54$. Replacing this value transforms the main expression into $|3 \operatorname{adj}(54 A^2)|$. Applying the rule for determinants of scalar multiples, this becomes $3^3 |\operatorname{adj}(54 A^2)|$, which simplifies to $27 |54 A^2|^2$. Substituting $|54 A^2| = 54^3 |A|^2 = 54^3 \times 4$, the expression evaluates to $27 \times (54^3 \times 4)^2$. Breaking down these terms yields $3^3 \times (2 \cdot 3^3)^6 \times 2^4 = 3^3 \times 2^6 \times 3^{18} \times 2^4 = 3^{21} \times 2^{10}$, which equals $3^{11} \cdot 3^{10} \cdot 2^{10} = 3^{11} \cdot 6^{10}$.

The matrix $B$ is formed by a similarity transformation $B = P A P^{-1}$, which means its $n$-th power simplifies to $B^n = P A^n P^{-1}$. Because

is a triangular matrix, its $n$-th power is simply

.

Substituting these into the expression for $B^n$ involves calculating

. Performing this multiplication step-by-step, the first product yields

, and completing the multiplication by the final matrix results in

.

Summing these matrices for $n=1$ to $50$ requires calculating the sum of each entry individually. The sum of the terms $\frac{n}{51}$ over this range is $\frac{1}{51} \times \frac{50 \times 51}{2} = 25$. Applying this result to the matrix components gives the sum of the top-left entries as $25 + 50 = 75$, the top-right as $25$, the bottom-left as $-25$, and the bottom-right as $50 - 25 = 25$. Adding these values together provides the final total of $75 + 25 - 25 + 25 = 100$.