Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations possesses infinitely many solutions when the determinant of the coefficient matrix is zero and the determinants formed by replacing columns with the constant vector also vanish, ensuring the equations are consistent and dependent. Evaluating the determinant of the coefficient matrix
$\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix}$
results in $\alpha(10-9) - 1(5-3) + 1(3-2) = \alpha - 1$, and setting this to zero requires $\alpha = 1$. Substituting $\alpha = 1$ and replacing the first column with the constants $5, 4, \beta$, the determinant
$\Delta_1 = \begin{vmatrix} 5 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 3 & 5 \end{vmatrix}$
expands to $5(1) - 1(20-3\beta) + 1(12-2\beta) = \beta - 3$. Setting $\Delta_1 = 0$ leads to $\beta = 3$, confirming that the ordered pair is $(\alpha, \beta) = (1, 3)$.

For a system of linear equations to have no solution, the coefficient matrix must be singular, meaning its determinant must be zero, and the system must be inconsistent. We begin by calculating the determinant of the coefficient matrix

. Expanding this determinant yields $-14a - 42$, which equals zero only when $a = -3$.

Substituting $a = -3$ into the system, we examine the linear dependencies between the equations to check for consistency. Applying the linear combination $3(\text{Eq1}) - 1(\text{Eq2}) - 2(\text{Eq3})$ to the system, the variable terms cancel out, leaving the constant expression $-3b + 1 = 0$. This implies that if $b = \frac{1}{3}$, the system would have infinite solutions; therefore, for the system to have no solution, we require $b \neq \frac{1}{3}$. Among the provided options, the pair that satisfies the condition $a = -3$ while resulting in an inconsistent system is $\left(-3, -\frac{1}{3}\right)$.

Determining the behavior of matrix powers often involves identifying a cycle, as finding the identity matrix simplifies the exponentiation process. Squaring $A$ results in

, and multiplying this by itself confirms that $A^4 = I$. Because $A^4=I$, the relation $A^n=A$ is equivalent to $A^{n-1}=I$, which holds when $n-1$ is a multiple of 4. Consequently, $n$ must follow the pattern $4k+1$ for non-negative integers. Within the range $n \in \{1, 2, \dots, 100\}$, the values follow the sequence $1, 5, 9, \dots, 97$. Since the largest term is $4(24)+1$, there are 25 such values.

When dealing with powers of matrices, it is helpful to first check if they possess the property of idempotency, where multiplying the matrix by itself returns the original matrix. Calculating the products for the given matrices reveals that $A^2 = A$ and $B^2 = B$, which means that for any positive integer exponent $k$, the matrices satisfy $A^k = A$ and $B^k = B$. Consequently, the expression $nA^n + mB^m$ simplifies directly to the sum $nA + mB$.

Substituting the given matrices into this expression yields

, which results in the matrix

. Setting this sum equal to the identity matrix

allows us to establish a system of equations by equating the corresponding components. From this, we obtain $2n - m = 1$ and $n - m = 0$, along with the secondary conditions $-2n + 2m = 0$ and $2m - n = 1$. The relations $n - m = 0$ and $-n + m = 0$ both imply that $n = m$. Substituting $n = m$ into the first equation $2n - m = 1$ gives $2n - n = 1$, which simplifies to $n = 1$ and therefore $m = 1$. As the pair $(1, 1)$ satisfies the original matrix equation and falls within the specified range of values, there is only one such pair in the set.

The Cayley-Hamilton theorem guarantees that every square matrix satisfies its own characteristic equation, which for any $2 \times 2$ matrix $A$ is expressed as
$\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$
This property implies that the matrix $A$ itself satisfies the corresponding matrix equation $A^2 - \text{tr}(A)A + \det(A)I = O$. Comparing this fundamental identity with the provided equation $A^2 + \gamma A + 18 I = O$, it is evident that the determinant of $A$ corresponds directly to the constant term of the equation. Consequently, $\det(A)$ is equal to 18.

The property of the matrix $M$ is captured by computing its square, which yields $M^2 = -\alpha^2 I$. Because $N$ is defined as the sum of even powers $M^{2k}$ from $k=1$ to $49$, it forms a geometric series containing 49 terms, where each term $M^{2k}$ is equivalent to $(-\alpha^2)^k I$. Applying the sum formula for a geometric progression with first term $-\alpha^2 I$ and common ratio $-\alpha^2$, we obtain $N = -\alpha^2 \left( \frac{1-(-\alpha^2)^{49}}{1-(-\alpha^2)} \right) I$, which simplifies to $N = -\alpha^2 \left( \frac{1+\alpha^{98}}{1+\alpha^2} \right) I$.

Since $I-M^2 = I - (-\alpha^2 I) = (1+\alpha^2) I$, the given equation $(I-M^2)N = -2I$ becomes $(1+\alpha^2) I \cdot \left( -\alpha^2 \frac{1+\alpha^{98}}{1+\alpha^2} \right) I = -2I$. Canceling the common factor $(1+\alpha^2)$ on the left side simplifies the expression to $-\alpha^2(1+\alpha^{98}) = -2$, or equivalently $\alpha^2(1+\alpha^{98}) = 2$. For a positive integer $\alpha$, the only value that satisfies this condition is $\alpha=1$.

When a system of three linear equations possesses infinitely many solutions, the determinant of the coefficient matrix must be equal to zero, and the determinants of the augmented matrices formed by the constants must also vanish. By constructing the determinant of coefficients from the equations $x+y+z=6$, $2x+5y+\alpha z=\beta$, and $x+2y+3z=14$, we set

equal to zero. Evaluating this determinant results in the expression $1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 8 - \alpha$, which directly implies that $\alpha$ must be $8$.

Substituting $\alpha=8$ back into the system, we can determine $\beta$ by setting the determinant of the matrix formed by replacing the first column with the constants-known as $\Delta_x$-equal to zero. This leads to the equation

. Expanding this determinant gives $6(15-16) - 1(3\beta-112) + 1(2\beta-70) = -6 - 3\beta + 112 + 2\beta - 70$, which simplifies to $36 - \beta = 0$, resulting in $\beta = 36$. Summing these values, $\alpha+\beta$ evaluates to $8+36$, which equals $44$.

A system of linear equations will either have no solution or infinitely many solutions when the determinant of its coefficient matrix is zero. Evaluating the determinant of the coefficients yields the equation $2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-1 - 1) = 0$, which simplifies to $7 - |\lambda| = 0$, confirming that $|\lambda| = 7$. When $\lambda = 7$, the third equation acts as a linear combination of the first two, specifically following the relation $P_3 = 5P_2 - 2P_1$, resulting in a consistent system with infinitely many solutions. Because the system with $\lambda = -7$ retains the same coefficients but results in a contradictory constant term of $-32$ instead of $24$, the equations cannot be satisfied simultaneously, making $\lambda = -7$ the value that leads to no solution.

The equality of two matrices requires that all their corresponding entries are identical, allowing us to determine the unknown scalars $\alpha$ and $\beta$ by equating them after evaluating the expression $\alpha A^2 + \beta A$. First, computing the square of the given matrix yields

, and substituting this into the equation produces the combined matrix

. Comparing the corresponding components creates the system $8\alpha = 2\beta$ and $-3\alpha + \beta = 2$, which uniquely determines the values $\alpha = 2$ and $\beta = 8$. Summing these two coefficients results in 10.

When the product of two non-zero 3 x 3 matrices results in the zero matrix $AB=0$, it implies that neither $A$ nor $B$ can be invertible, as the existence of an inverse for either would force the other to be the zero matrix. Since $A$ is not invertible, its determinant $|A|$ must be zero, which indicates that the homogeneous system $AX=0$ cannot have a unique solution. Consequently, such a system must possess non-trivial solutions, which inevitably leads to the existence of infinitely many solutions.

A system of linear equations is consistently defined with a unique solution provided the determinant of the coefficient matrix is non-zero. When the determinant equals zero, the system faces the possibility of being inconsistent or having infinitely many solutions, requiring us to isolate the specific values of $k$ that trigger this condition.

To identify these critical values, calculate the determinant of the matrix formed by the coefficients:
$\begin{vmatrix} -k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3 \end{vmatrix} = -k(12 + k) - 3(-45 - 4k) - 14(-15 + 16)$
Simplifying this expression yields $-12k - k^2 + 135 + 12k - 14$, which reduces to $121 - k^2$. Equating the determinant to zero results in $121 - k^2 = 0$, meaning the determinant vanishes only when $k = 11$ or $k = -11$. Consequently, the system remains consistent for all real values of $k$ excluding these two specific points.

Elementary row operations consist of interchanging two rows, multiplying a row by a non-zero scalar, or adding a scalar multiple of one row to another. By performing these specific transformations on the matrix

, one can test the validity of each option in a single step. The operation $R_1 \rightarrow R_1 + R_2$ produces

, while the swap operation $R_1 \leftrightarrow R_2$ results in

. Furthermore, applying $R_2 \rightarrow R_2 + 2R_1$ updates the second row by adding twice the first, yielding

. Because the matrix

cannot be formed through any of these single-step transformations, it remains the only unreachable result among the choices.

The condition $A=A^{-1}$ is equivalent to $A^2=I$. By computing the product of matrix

with itself, we obtain:
$A^2 = \begin{pmatrix} a^{2+}bc & b(a+d) \\ c(a+d) & d^{2+}bc \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
This matrix equation leads to a system of equations: $a^{2+}bc=1$, $d^{2+}bc=1$, $b(a+d)=0$, and $c(a+d)=0$. From the last two equations, either $a+d \neq 0$, which implies $b=0$ and $c=0$, or $a+d=0$, which implies $d=-a$.

In the first case, where $a+d \neq 0$, we have $b=c=0$. The conditions $a^{2+}bc=1$ and $d^{2+}bc=1$ simplify to $a^2=1$ and $d^2=1$. This gives two possible matrices: $A=I$ where $a=d=1$, and $A=-I$ where $a=d=-1$. Both satisfy the condition $a+d \neq 0$ and the required constraints, contributing two matrices to the total count.

In the second case, where $d=-a$, the equations $a^{2+}bc=1$ and $d^{2+}bc=1$ are identical, both reducing to $bc=1-a^2$. We consider the possible values of $a \in \{-1, 0, 1, 2, \dots, 10\}$ that satisfy this. When $a=0$, we have $bc=1$, which for the given set $S=\{-1, 0, 1, 2, \dots, 10\}$ only occurs for pairs $(b,c)=(1,1)$ and $(-1,-1)$, providing two matrices. When $a=1$, the condition $bc=1-1^2=0$ implies that either $b=0$ or $c=0$. With 12 choices for $b$ and 12 for $c$, accounting for the overlap of the pair $(0,0)$, we find $12+12-1=23$ possible pairs. Similarly, when $a=-1$, the condition $bc=0$ again yields 23 possible pairs. Summing these results gives $2+2+23+23=50$, which is the total number of such matrices.

A system of two linear equations possesses infinitely many solutions when the ratios of corresponding coefficients and constant terms are identical, which requires satisfying the condition
$\frac{2}{\alpha} = \frac{-3}{5} = \frac{\gamma + 5}{\beta + 1}$
From the equality involving $\alpha$, cross-multiplying gives $\alpha = -\frac{10}{3}$, while the equality between the latter two ratios leads to the relation $-3(\beta + 1) = 5(\gamma + 5)$, which simplifies to $3\beta + 5\gamma = -28$. By substituting these findings into the expression $|9\alpha + 3\beta + 5\gamma|$, we compute $|9(-\frac{10}{3}) - 28|$, which equates to $|-30 - 28|$ and results in a final value of $58$.

A system of linear equations fails to provide a unique solution when the determinant of the coefficient matrix is equal to zero, rendering the system inconsistent. To identify the values of $\lambda$ that result in no solution, we evaluate the determinant of the coefficient matrix constructed from the system:

Expanding this determinant across the first row gives $\Delta = 2(3(\lambda^2 - |\lambda|) - 1) + 3(1(\lambda^2 - |\lambda|) + 3) + 5(-1 - 9)$. Simplifying this expression leads to $\Delta = 9\lambda^2 - 9|\lambda| - 43$.

Setting the determinant equal to zero to satisfy the condition for inconsistency results in the quadratic equation $9|\lambda|^2 - 9|\lambda| - 43 = 0$. Using the quadratic formula to solve for $|\lambda|$, we obtain two distinct roots, one of which is positive and one of which is negative. Since the absolute value $|\lambda|$ must inherently be non-negative, we discard the negative root. The remaining positive value for $|\lambda|$ yields two valid real values for $\lambda$, specifically the positive and negative versions of the square root of that constant. Consequently, there are two distinct real values of $\lambda$ for which the system has no solutions.

A system of linear equations fails to possess a unique solution when the determinant of its coefficient matrix equals zero. The condition for the given system to have no solution requires setting the determinant of the coefficient matrix to zero:

$\begin{vmatrix} 3\sin 3\theta & -1 & 1 \\ 3\cos 2\theta & 4 & 3 \\ 6 & 7 & 7 \end{vmatrix} = 0$

Expanding this determinant along the first row gives the expression $3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$. Simplifying these terms results in $21\sin 3\theta + 42\cos 2\theta - 42 = 0$. Dividing the entire equation by 21 leads to the simplified trigonometric equation $\sin 3\theta + 2\cos 2\theta - 2 = 0$. Using the identities $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$, the equation becomes $3\sin\theta - 4\sin^3\theta + 2(1 - 2\sin^2\theta) - 2 = 0$. After algebraic reduction, this simplifies to $4\sin^3\theta + 4\sin^2\theta - 3\sin\theta = 0$, which can be factored as $\sin\theta(2\sin\theta - 1)(2\sin\theta + 3) = 0$.

The product is zero when $\sin\theta = 0$, $\sin\theta = 1/2$, or $\sin\theta = -3/2$. Because the value of sine must remain within the range $[-1, 1]$, the case $\sin\theta = -3/2$ is impossible. Within the interval $(0, 4\pi)$, the equation $\sin\theta = 0$ yields values of $\theta$ at $\pi$, $2\pi$, and $3\pi$. Similarly, the equation $\sin\theta = 1/2$ yields values at $\theta = \pi/6$, $5\pi/6$, $13\pi/6$, and $17\pi/6$. Summing these results gives a total of 7 possible values for $\theta$.

Identifying the recursive behavior in the powers of matrix $A$ provides the most efficient path to the solution. Computing the square of the matrix yields:
$A^2 = \begin{bmatrix}-1&2&3\\0&1&6\\0&0&-1\end{bmatrix} \begin{bmatrix}-1&2&3\\0&1&6\\0&0&-1\end{bmatrix} = \begin{bmatrix}1&0&6\\0&1&0\\0&0&1\end{bmatrix}$
This progression establishes that for any even integer $k$, the matrix power follows the form

. Substituting this into the equation $X^{\prime} A^{k} X = 33$ with

allows us to perform the matrix multiplication:
$\begin{bmatrix}1&1&1\end{bmatrix} \begin{bmatrix}1&0&3k\\0&1&0\\0&0&1\end{bmatrix} \begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}1&1&3k+1\end{bmatrix} \begin{bmatrix}1\\1\\1\end{bmatrix} = 3k+3 = 33$
Solving the resulting linear equation $3k = 30$ confirms that $k = 10$.

A system of linear equations possesses infinitely many solutions when the determinant of its coefficient matrix vanishes, indicating that the equations are linearly dependent and consistent. Computing the determinant for the provided system yields the expression

$\begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix} = 2(-3\delta - 8) - 1(\delta - 2) - 1(4 + 3) = -7\delta - 21$

Setting this determinant to zero results in $\delta = -3$.

For the system to maintain infinite solutions, the determinant of the matrix formed by replacing the first column with the constants must also equate to zero, ensuring the equations are consistent:

$\begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{vmatrix} = 7(9 - 8) - 1(-3 - 2k) - 1(4 + 3k) = 6 - k$

Requiring $6 - k = 0$ provides $k = 6$. Consequently, the sum $\delta + k$ evaluates to $-3 + 6 = 3$.

For a $3 \times 3$ invertible matrix $A$, the determinant of an adjoint matrix follows the identity $|\operatorname{adj} M| = |M|^{n-1}$, which simplifies here to $|\operatorname{adj} M| = |M|^2$. Additionally, scaling a matrix by a factor $k$ impacts its determinant according to $|k M| = k^n |M|$, which in this case is $|k M| = k^3 |M|$. Applying these properties to the given equation $|\operatorname{adj}(24 A)| = |\operatorname{adj}(3 \operatorname{adj}(2 A))|$, we can equate the determinant of both sides as $|24 A|^2 = |3 \operatorname{adj}(2 A)|^2$.

Expanding both sides, the left expression becomes $(24^3 |A|)^2$, which simplifies to $24^6 |A|^2$. On the right side, the determinant is $(3^3 |\operatorname{adj}(2 A)|)^2$, or $3^6 |\operatorname{adj}(2 A)|^2$. Using the relation $|\operatorname{adj}(2 A)| = |2 A|^2$, and knowing $|2 A| = 2^3 |A| = 8 |A|$, the term $|\operatorname{adj}(2 A)|^2$ becomes $(8^2 |A|^2)^2 = 8^4 |A|^4$. Substituting these values gives:
$24^6 |A|^2 = 3^6 \cdot 8^4 \cdot |A|^4$
Because $24^6 = (3 \cdot 8)^6 = 3^6 \cdot 8^6$, we can rewrite the equation as $3^6 \cdot 8^6 \cdot |A|^2 = 3^6 \cdot 8^4 \cdot |A|^4$. Dividing both sides by $3^6 \cdot 8^4 \cdot |A|^2$ leaves $8^2 = |A|^2$, which evaluates to $64$, or $2^6$.

The collinearity of the points $(\alpha, 1)$, $(1, \alpha)$, and $(1, -1)$ requires that the determinant of the matrix formed by these coordinates is zero, as the area of the triangle they span must be null. To identify the unique solution $(x^*, y^*, z^*)$ of the system $x+y+\alpha z=2$, $3x+y+z=4$, and $x+2z=1$, we express $x = 1-2z$ from the third equation and substitute it into the second to find $y = 1+5z$. Inserting these into the first equation produces $z(\alpha+3)=0$, which confirms that a unique solution exists only when $\alpha \neq -3$, leading to $z=0$ and the coordinate values $x^*=1$ and $y^*=1$.

With these values, the collinearity condition is expressed as

. Expanding this determinant yields $\alpha(\alpha+1) - 1(1-1) + 1(-1-\alpha) = 0$, which simplifies to the equation $\alpha^2 - 1 = 0$. Since $\alpha = 1$ and $\alpha = -1$ both satisfy the uniqueness constraint $\alpha \neq -3$, the sum of their absolute values is $|1| + |-1| = 2$.