Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations lacks a solution when the determinant of its coefficient matrix equals zero, indicating that the constraints are either inconsistent or linearly dependent. Setting the determinant of the coefficient matrix for the given equations leads to the expansion
$\begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$
Expanding this expression produces $\beta - 4\alpha - 4\beta + 16\alpha = 0$. Combining the terms results in $12\alpha - 3\beta = 0$, which implies $3\beta = 12\alpha$. Dividing by $\alpha$, the ratio $\frac{\beta}{\alpha}$ resolves to 4.

A system of linear equations $AX=B$ possesses no solution if the determinant of the coefficient matrix $|A|$ equals zero, and the rank of the augmented matrix $[A|B]$ is greater than the rank of the coefficient matrix $A$. Setting the determinant of the coefficient matrix to zero provides the following calculation:

$|A| = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 3(\alpha + 2) - 1(2 + 1) + 4(4 - \alpha) = 0$

The expansion $3\alpha + 6 - 3 + 16 - 4\alpha = 0$ simplifies to $-\alpha + 19 = 0$, resulting in $\alpha = 19$. Testing for consistency with $\alpha=19$ by substituting $x = 4 - 2y - z$ from the third equation into the other two equations leads to:

$3(4 - 2y - z) + y + 4z = 3 \implies -5y + z = -9$

$2(4 - 2y - z) + 19y - z = -3 \implies 15y - 3z = -11$

Scaling the first derived equation by $-3$ produces $15y - 3z = 27$, which directly contradicts the second derived equation $15y - 3z = -11$, confirming that no solution exists.

A system of linear equations possesses a unique solution if and only if the determinant of its coefficient matrix is non-zero. The determinant of the system's coefficient matrix is given by

Expanding this determinant along the first column results in $1((n-2) - (n^{2-}1)) - 1(-n - (n-1)) + 0$, which simplifies to $n - 2 - n^2 + 1 + n + n - 1 = -n^2 + 3n - 2$. Setting the determinant to be non-zero gives $-(n-1)(n-2) \neq 0$, which implies $n \neq 1$ and $n \neq 2$. Given that $n$ is the outcome of a fair die roll, $n \in \{1, 2, 3, 4, 5, 6\}$, the values of $n$ for which a unique solution exists are $\{3, 4, 5, 6\}$. There are 4 such values, meaning the probability is $\frac{4}{6}$, identifying $k=4$. The sum of $k$ and all possible values of $n$ is $4 + 3 + 4 + 5 + 6 = 22$.

The matrix $B = (I+A)^{-1}$ combined with $A = I-C$ simplifies to $B = (2I-C)^{-1}$. Since any matrix $C$ commutes with a polynomial or inverse function of itself, $C(2I-C)^{-1} = (2I-C)^{-1}C$, which implies $CB = BC$. Given $BC = \begin{pmatrix} 1 & -5 \\ -1 & 2 \end{pmatrix}$, the system $CB \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \end{pmatrix}$ implies

. The resulting equations are $x_1 - 5x_2 = 12$ and $-x_1 + 2x_2 = -6$, and adding these equations yields $-3x_2 = 6$, so $x_2 = -2$. Substituting this value into $x_1 - 5x_2 = 12$ gives $x_1 + 10 = 12$, resulting in $x_1 = 2$, so the sum $x_1 + x_2$ is 0.

Isolating unknown matrices in equations like $PA=B$ and $AQ=B$ requires the use of the inverse matrix, $A^{-1}$, while strictly respecting the order of multiplication since matrix algebra is non-commutative. Once the inverse is found, the matrices $P$ and $Q$ can be calculated directly, allowing for the trace of $2(P+Q)$ to be determined.

For the matrix

, the determinant is $(2)(-2) - (-2)(4) = 4$. Using the standard formula for the inverse of a $2 \times 2$ matrix, we obtain:
$A^{-1} = \frac{1}{4} \begin{bmatrix} -2 & 2 \\ -4 & 2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}$
With the inverse established, we calculate $P$ by right-multiplying $B$ by $A^{-1}$, which gives

. Conversely, calculating $Q$ involves left-multiplying $B$ by $A^{-1}$, resulting in

.

The trace of a matrix is the sum of its diagonal elements. Given the linearity of the trace operator, the trace of $2(P+Q)$ is equivalent to $2(\text{tr}(P) + \text{tr}(Q))$. Substituting the values derived from our matrix products, the sum is $(-21) + 4 + (-2) + (-15) = -34$. Taking the absolute value of this result yields 34.

For square matrices $Q = D_1 P D_2$ where $D_1 = \operatorname{diag}(a_1, a_2, a_3)$ and $D_2 = \operatorname{diag}(b_1, b_2, b_3)$, the determinant follows the relation $\det(Q) = (\prod_{i=1}^3 a_i) (\prod_{j=1}^3 b_j) \det(P)$. Given $q_{ij} = 2^i 2^{j-1} p_{ij}$, the row factors are $a_i = 2^i$ and column factors are $b_j = 2^{j-1}$. The product of the row factors is $2^{1+2+3} = 2^6$ and the product of the column factors is $2^{0+1+2} = 2^3$. Substituting these values yields $2^{10} = 2^6 \cdot 2^3 \cdot \det(P)$, which simplifies to $2^{10} = 2^9 \det(P)$, resulting in $\det(P) = 2$. The determinant of the adjoint of the adjoint is given by the general formula $\det(\operatorname{adj}(\operatorname{adj} P)) = (\det P)^{(n-1)^2}$ for a matrix of order $n$. With $n=3$, this expression becomes $(\det P)^{(3-1)^2} = (\det P)^4$. Substituting $\det(P) = 2$ gives $2^4 = 16$.

The power of the matrix

follows a general pattern defined by

. Calculating this for $n=15$ yields

.

Adding the given matrix

to $A^{15}$ results in the sum:
$A^{15}+B = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$
The matrix equation

reduces to the linear relation $2x - 11y = 0$, or $2x = 11y$. Among the provided options, substituting $x=11$ and $y=2$ satisfies this equality.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic equation, which for

is $(\lambda-1)^2=0$, implying $(A-I)^2=O$. Applying the binomial expansion to $A^{100} = (I + (A-I))^{100}$ yields
$A^{100} = I + 100(A-I)$
since $(A-I)^k = O$ for all $k \ge 2$. Comparing this result with the given equation $A^{100} = 100B + I$ reveals

. Evaluating the square of this matrix gives
$B^2 = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
which confirms that $B$ is a nilpotent matrix. Because $B^2 = O$, it follows that $B^{100} = B^{98} \cdot B^2 = O$, and therefore the sum of all its elements is 0.

A system of linear equations is consistent with infinitely many solutions when the rank of the augmented matrix is less than the number of variables, a condition satisfied when the final row of the row-echelon form becomes all zeros. Constructing the augmented matrix for the system and performing the row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ yields the following matrix:
$\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 3 & a-2 & 24 \\ 0 & 1 & 2 & b-6 \end{bmatrix}$
Interchanging $R_2$ and $R_3$ followed by the operation $R_3 \to R_3 - 3R_2$ transforms the system into the reduced form
$\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & b-6 \\ 0 & 0 & a-8 & 42-3b \end{bmatrix}$
For the system to possess infinitely many solutions, the last row must represent the identity $0=0$, which necessitates that both $a-8=0$ and $42-3b=0$. Solving these equations leads to $a=8$ and $b=14$.

A system of homogeneous linear equations admits a non-trivial solution if and only if the determinant of its coefficient matrix equals zero. Setting the determinant of the coefficients to zero yields:

Expanding this determinant results in $6\cos 3\theta + 12\cos 2\theta + 12 = 0$, which simplifies to $\cos 3\theta + 2\cos 2\theta + 2 = 0$. Substituting the standard trigonometric identities $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ and $\cos 2\theta = 2\cos^2 \theta - 1$ leads to the equation $(4\cos^3 \theta - 3\cos \theta) + 2(2\cos^2 \theta - 1) + 2 = 0$. This reduces to $4\cos^3 \theta + 4\cos^2 \theta - 3\cos \theta = 0$, which factors into $\cos \theta (2\cos \theta + 3)(2\cos \theta - 1) = 0$. Since the value of $\cos \theta$ must fall within the range $[-1, 1]$, the factor $(2\cos \theta + 3)$ cannot equal zero, leaving $\cos \theta = 0$ and $\cos \theta = \frac{1}{2}$ as the only valid solutions. In the interval $[0, 2\pi]$, $\cos \theta = 0$ yields $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$, while $\cos \theta = \frac{1}{2}$ yields $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. Calculating the sum of these angles gives $\frac{\pi}{2} + \frac{3\pi}{2} + \frac{\pi}{3} + \frac{5\pi}{3} = 2\pi + 2\pi = 4\pi$.

The characteristic equation of a $2 \times 2$ matrix $M$ is $\lambda^2 - \text{tr}(M)\lambda + \det(M) = 0$. Comparing this with the given equations $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$ allows us to determine the values of $\alpha$ and $\beta$. For matrix $A$, the trace $1+\alpha$ equals $4$, so $\alpha=3$. For matrix $B$, the determinant $6-3\beta$ equals $-6$, which leads to $\beta=4$.

Using these values,

and

, we find

and

. Multiplying these results gives

. Because the transpose of this product is

, which does not equal the matrix provided in the statement, (S1) is incorrect.

For statement (S2), the property $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$ for an $n \times n$ matrix applies, simplifying to $\det(\operatorname{adj}(M)) = \det(M)$ for a $2 \times 2$ matrix. Calculating

, we find its determinant is $(4)(5) - (5)(5) = -5$. Since $\det(A+B) = -5$, it follows that $\det(\operatorname{adj}(A+B)) = -5$, which confirms statement (S2) is correct.

Since $A+A^T=O$, $A$ is a skew-symmetric matrix of the form

, and from the product

, we find $a=-3$ and $c-b=2$. Given

, it follows that $2b=6 \implies b=3$ and $2c=10 \implies c=5$, which yields

.

Using the properties $\det(\operatorname{adj}(M)) = (\det M)^{n-1}$ and $\det(kM) = k^n \det M$ for a $3 \times 3$ matrix $M$, the required determinant is $\det(\operatorname{adj}(2 \operatorname{adj}(A+I))) = (\det(2 \operatorname{adj}(A+I)))^2$. Since $\det(2 \operatorname{adj}(A+I)) = 2^3 \det(\operatorname{adj}(A+I)) = 8 (\det(A+I))^2 = 8(44)^2 = 8(2^4 \cdot 11^2) = 2^7 \cdot 11^2$, the final expression is $(2^7 \cdot 11^2)^2 = 2^{14} \cdot 11^4$. Comparing this to $2^\alpha \cdot 3^\beta \cdot 11^\gamma$, we identify $\alpha=14, \beta=0, \gamma=4$, so $\alpha+\beta+\gamma = 14+0+4 = 18$.

The trace of $A^T A$ is equivalent to the sum of the squares of all elements in the matrix $A$. Given that $A$ is a $3 \times 2$ matrix, it contains 6 elements $a_{ij}$ chosen from the set $\{-2, -1, 0, 1, 2\}$, where the squares of these elements are $a_{ij}^2 \in \{0, 1, 4\}$. The condition that the sum of the diagonal elements of $A^T A$ is 5 implies that the sum of the squares of the six elements in $A$ must be 5.

This condition is satisfied by two distinct combinations of squared values: $\{4, 1, 0, 0, 0, 0\}$ and $\{1, 1, 1, 1, 1, 0\}$. For the first combination, there are $\binom{6}{1}$ positions to place the element with square 4 and $\binom{5}{1}$ positions to place the element with square 1. Since square 4 is produced by $a_{ij} \in \{2, -2\}$ (2 choices) and square 1 is produced by $a_{ij} \in \{1, -1\}$ (2 choices), the total number of matrices for this case is $6 \times 5 \times 2 \times 2 = 120$.

For the second combination, there are $\binom{6}{1}$ ways to choose the position for the element with square 0, and the remaining 5 elements must each correspond to a square of 1, which has 2 choices ($\pm 1$). This yields $6 \times 2^5 = 6 \times 32 = 192$ matrices. Combining both cases results in $120 + 192 = 312$ total possible matrices.

The determinant on the left side expands to $1 - \cos^2\alpha - \cos^2\beta - \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma$, while the determinant on the right is $2\cos\alpha\cos\beta\cos\gamma$. Equating these two expressions yields $1 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) = 0$, which simplifies to $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. Since this result contradicts the value of $\frac{3}{2}$ given in the first statement, statement I is false.

The second determinant simplifies to the linear polynomial $24x - 12$ by applying column operations such as $C_2 \to C_2 - C_3$ and evaluating the matrix

. Comparing this to $px + q$, we identify $p = 24$ and $q = -12$, which results in $p^2 = 576$ and $196q^2 = 196 \times 144 = 28224$. Since $576 \neq 28224$, the condition $p^2 = 196q^2$ is not met, so statement II is also false.

Any matrix $A$ satisfying the characteristic equation $A^2 - 4A + 3I = 0$ must have eigenvalues that are the roots of $\lambda^2 - 4\lambda + 3 = 0$. Factoring this quadratic equation yields eigenvalues 1 and 3, which implies the matrix has a trace of 4 and a determinant of 3. We seek the number of matrices with elements in $\{0, 1, 2, 3, 4\}$ that satisfy these conditions, specifically $a + d = 4$ and $ad - bc = 3$.

Considering the possible pairs $(a, d)$ that satisfy $a + d = 4$, we identify the valid cases as $(1, 3)$, $(2, 2)$, and $(3, 1)$. For the pair $a = 1$ and $d = 3$, the determinant condition $ad - bc = 3$ becomes $3 - bc = 3$, which simplifies to $bc = 0$. This product is satisfied by 9 pairs of $(b, c)$, namely $(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (2, 0), (3, 0), (4, 0)$. Similarly, the case $a = 3$ and $d = 1$ yields 9 additional matrices. For $a = 2$ and $d = 2$, the condition $ad - bc = 3$ requires $4 - bc = 3$, meaning $bc = 1$, which is only possible when $b = 1$ and $c = 1$. Summing these possibilities, the total number of valid matrices is $9 + 9 + 1 = 19$.

For a system of linear equations to possess infinitely many solutions, the coefficient matrix must be singular, which means its determinant must equal zero, and the overall system must remain consistent. Calculating the determinant of the coefficient matrix

leads to $1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = \lambda - 5$. Setting this expression to zero identifies the value of $\lambda$ as $5$.

For the system to be consistent, the determinant of the augmented matrix formed by replacing the third column with the constant vector must also vanish. Evaluating

results in $1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2) = 0$, which simplifies to $\mu - 13 = 0$, giving $\mu = 13$. Combining these results leads to $\lambda + \mu = 5 + 13 = 18$.

The solution to the system $AX = B$ is given by $X = A^{-1}B$, which can be expressed using the adjoint of $A$ as $X = \frac{1}{|A|} (\operatorname{adj} A) B$. The determinant of the adjoint matrix is related to the determinant of $A$ by the formula $|\operatorname{adj} A| = |A|^{n-1}$, where $n$ is the order of the matrix. For the given $3 \times 3$ matrix, this becomes $|\operatorname{adj} A| = |A|^2$. Calculating the determinant of the given adjoint matrix:
$|\operatorname{adj} A| = 4(0 - (-10)) - 2(-15 - 5) + 2(10 - 0) = 4(10) - 2(-20) + 2(10) = 40 + 40 + 20 = 100$.
Thus, $|A|^2 = 100$, which implies $|A| = \pm 10$.

To find the solution vector $X$, we multiply the adjoint matrix by the vector $B$:

.
The solution

is therefore