Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations lacks a unique solution when the determinant of the coefficient matrix, $D$, is zero. Inconsistency specifically arises if $D = 0$ while at least one of the related determinants $D_1, D_2,$ or $D_3$ remains non-zero. Calculating the determinant of the coefficient matrix yields:
$D = \begin{vmatrix} 2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4 \end{vmatrix} = 2(3\lambda + 2)(\lambda - 3)$
Setting this expression to zero reveals that the system could only potentially be inconsistent or have infinite solutions when $\lambda = 3$ or $\lambda = -\frac{2}{3}$.

Evaluating the other determinants at $\lambda = 3$ results in $D_1 = -2(\lambda - 3)$, $D_2 = -2(\lambda + 1)(\lambda - 3)$, and $D_3 = -2(\lambda - 3)$ all vanishing to zero, which indicates the existence of infinitely many solutions rather than inconsistency. When testing $\lambda = -\frac{2}{3}$, however, the determinant $D$ is zero while $D_1, D_2,$ and $D_3$ are all non-zero. This identifies that the system is inconsistent for exactly one negative value of $\lambda$.

The set of all $2 \times 2$ matrices composed of entries from $\{0, 1\}$ that possess a non-zero determinant is finite, which allows for direct evaluation of the given statements by testing the behavior of these specific candidates. This constrained environment means we can identify counterexamples for falsifiable claims or verify that properties hold across all valid instances for true claims.

Regarding statement (P), we evaluate whether $A \neq I_{2}$ necessitates that $\left| A \right| = -1$. We can identify the matrix

, which is clearly not the identity matrix. Calculating its determinant results in $1(1) - 1(0) = 1$, which is not $-1$. Because this matrix satisfies the requirement of having a non-zero determinant but violates the conclusion of the statement, (P) is false.

For statement (Q), we examine all valid matrices with entries $\{0, 1\}$ that satisfy $\left| A \right| = 1$. The matrices meeting this condition are the identity matrix

and the matrices

and

. In every instance, the sum of the diagonal entries, $\operatorname{tr}(A)$, is $1 + 1 = 2$. Since this property holds for all matrices where the determinant is $1$, (Q) is true.

Three planes intersect along a single line only if their defining system of linear equations is dependent, meaning there must be infinitely many solutions. This condition forces the determinant of the coefficient matrix to be zero, calculated as

, which simplifies to $3\alpha + 9 = 0$ and leads to $\alpha = -3$.

For the system to remain consistent while having infinitely many solutions, the determinant formed by replacing the coefficients of $z$ with the constant vector must also equal zero:
$\begin{vmatrix} 1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5 \end{vmatrix} = 0$
Computing this determinant gives $1(35 - 5\beta) - 4(5 - \beta) + 1(5 - 7) = 13 - \beta = 0$, resulting in $\beta = 13$. Combining these two values yields a sum of 10.

A homogeneous system of linear equations possesses non-zero solutions if and only if the determinant of its coefficient matrix equals zero. Setting the determinant of the matrix formed by the system's coefficients to zero gives:

Applying the row transformations $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ simplifies the determinant to:

Factoring $(\lambda-3)$ out of the first and second rows results in $(\lambda-3)^{2}$ multiplied by the remaining determinant

. Evaluating this smaller determinant yields $(3\lambda+1) + (3\lambda-1)$, which simplifies to $6\lambda$. The resulting equation is $6\lambda(\lambda-3)^{2}=0$, providing the distinct values $\lambda=0$ and $\lambda=3$. These values sum to 3.

Applying elementary row operations is the most efficient way to evaluate this determinant because it simplifies the matrix before expansion. Subtracting the second row from the first results in a new first row of $(-1, 1, 0)$, and subtracting the third row from the second yields a new second row of $(1, 0, -1)$. Expanding this simplified determinant along the first row gives the expression $-1(\sin^2x) - 1(1 + \sin2x + \cos^2x)$, which simplifies to $-\sin^2x - \cos^2x - 1 - \sin2x$, further reducing to $-2 - \sin2x$. Since the value of $\sin2x$ oscillates between $-1$ and $1$, the entire expression ranges from a minimum of $-1 - 2 = -3$ to a maximum of $1 - 2 = -1$.

Simplifying this determinant relies on elementary column operations, beginning with $C_3 \rightarrow C_3 - C_1$, which transforms the third column into the elements $a, b, c$. Applying $C_2 \rightarrow C_2 - C_3$ then reduces the second column to $y, y, y$ because $(a+y)-a=y$, $(b+y)-b=y$, and $(c+y)-c=y$. Factoring $y$ out of this second column creates a simpler structure that is effectively reduced by row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$.

Expanding the resulting determinant along the second column yields an expression involving the terms $(y-x)(c-a) - (z-x)(b-a)$. By utilizing the given constraint $a+x=b+y=c+z+1$, we substitute $y-x=a-b$ and $z-x=a-c-1$ into this expression. Completing the algebraic reduction confirms that the determinant simplifies to $y(a-b)$.

Matrix exponentiation allows for the determination of specific entries by building up from the square of the given matrix. Computing the square of

yields the product

. Squaring this result to find $A^4$ reveals that the first entry, $a_{11}$, is equal to $(x^{2+}1)^2 + x^2$.

Setting this expression equal to 109 results in the equation $(x^{2+}1)^2 + x^2 = 109$. Expanding the squared term leads to $x^4 + 2x^2 + 1 + x^2 = 109$, which simplifies to $x^4 + 3x^2 - 108 = 0$. Factoring this quadratic expression in terms of $x^2$ as $(x^2 + 12)(x^2 - 9) = 0$ confirms that $x^2 = 9$, since the square of a real number cannot be negative. Because the element $a_{22}$ in the calculated $A^4$ matrix is $x^2 + 1$, substituting the value 9 yields $9 + 1$, resulting in 10.

The pattern of powers for a matrix of this form,

, follows a predictable progression where

. Given $n=5$ and $\theta = \frac{\pi}{24}$, we identify the elements $a = \cos 5\theta$, $b = i\sin 5\theta$, $c = i\sin 5\theta$, and $d = \cos 5\theta$.

Substituting these into the provided relations, $a^2 + b^2$ evaluates to $\cos^2 5\theta + (i\sin 5\theta)^2$, which simplifies to $\cos^2 5\theta - \sin^2 5\theta = \cos 10\theta$. Since $10\theta = \frac{10\pi}{24} = 75^{\circ}$, this value lies between 0 and 1, confirming the truth of the first statement. Evaluating $a^2 - d^2$ yields $\cos^2 5\theta - \cos^2 5\theta = 0$, which is also true. Calculating $a^2 - b^2$ gives $\cos^2 5\theta - (i\sin 5\theta)^2 = \cos^2 5\theta + \sin^2 5\theta = 1$, indicating that the claim setting this value equal to $\frac{1}{2}$ is incorrect. Finally, $a^2 - c^2$ follows the same logic as the previous calculation, simplifying to $\cos^2 5\theta + \sin^2 5\theta = 1$, which is a true statement.

The determinants of an $n \times n$ matrix $M$ relate to its adjoint and scalar multiples through the identities $|\operatorname{adj} M| = |M|^{n-1}$ and $|kM| = k^n|M|$. First, calculating the determinant of the given $3 \times 3$ matrix $A$ yields $|A| = 6$. Since $B$ is defined as the adjoint of $A$, the numerator $|\operatorname{adj} B|$ becomes $|\operatorname{adj}(\operatorname{adj} A)|$, which simplifies using the property for adjoints of adjoints to $|A|^{(3-1)^2}$, or $|A|^4$. For the denominator, $|C| = |3A|$ evaluates to $3^3|A|$ since the matrix is of order 3, which equals $27|A|$. Combining these properties, the expression simplifies as follows:

$\frac{|\operatorname{adj} B|}{|C|} = \frac{|A|^4}{27|A|} = \frac{|A|^3}{27}$

Substituting the value $|A| = 6$ into this result gives $\frac{6^3}{27} = \frac{216}{27}$, which simplifies to 8.

The determinant of the adjoint of a square matrix $A$ of order $n$ relates to the original determinant through the property $|\text{adj} A| = |A|^{n-1}$. Applying this to the $3 \times 3$ matrix $A$, we find $|\text{adj} A| = |A|^2$. Computing the determinant of the given adjoint matrix

via expansion along the first row gives $2(0 - (-4)) - (-1)(1 - 2) + 1(2 - 0) = 8 - 1 + 2 = 9$. Thus, $|A|^2 = 9$ implies $|A| = \pm 3$, making the absolute value $|\lambda| = 3$.

Since $B = \text{adj}(\text{adj} A)$ can be viewed as the adjoint of the matrix $C = \text{adj} A$, the determinant $|B|$ is equal to $|\text{adj} C| = |C|^{3-1} = |C|^2 = 9^2 = 81$. Using the properties of determinants, where $|(B^{-1})^{T}| = |B^{-1}| = \frac{1}{|B|}$, the value of $\mu$ becomes $\frac{1}{81}$. This results in the ordered pair $(|\lambda|, \mu)$ being $\left(3, \frac{1}{81}\right)$.

A system of linear equations possesses more than two solutions only when it has infinitely many solutions, a condition occurring strictly when the determinant of the coefficient matrix equals zero. Applying this constraint requires evaluating the following determinant:

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 2 & \lambda \end{vmatrix} = 0$

Expanding this expression results in $1(2\lambda - 6) - 1(\lambda - 9) + 1(2 - 6) = 0$, which simplifies to $\lambda - 1 = 0$, establishing that $\lambda = 1$. For the system to maintain consistency while having infinitely many solutions, the determinant of the matrix formed by replacing the first column of the coefficient matrix with the constant terms must also equal zero:

$\begin{vmatrix} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \mu & 2 & 1 \end{vmatrix} = 0$

Calculating this determinant yields $6(2 - 6) - 1(10 - 3\mu) + 1(20 - 2\mu) = -24 - 10 + 3\mu + 20 - 2\mu = \mu - 14 = 0$, confirming that $\mu = 14$. Substituting these values into the required expression provides $14 - 1^2$, resulting in a final value of 13.

A system of linear equations is inconsistent when the equations contradict one another, appearing mathematically as an impossible equality such as $0 = k$ for a non-zero constant $k$. Applying the operation $2 \times (\text{ii}) - 2 \times (\text{i}) - (\text{iii})$ allows us to combine the equations into the form $0 = 2\mu - 2 - \delta$, which dictates that the system is consistent only when $\delta = 2(\mu - 1)$. Any ordered pair $(\mu, \delta)$ that fails to satisfy this equality creates a contradiction, rendering the system inconsistent. Inserting the values from the pair $(4, 3)$ into the derived expression yields $0 = 2(4) - 2 - 3$, which simplifies to $0 = 3$, demonstrating the necessary contradiction.

Any system of homogeneous linear equations always permits the trivial solution, but the existence of non-trivial solutions depends on whether the determinant of the coefficient matrix is zero. Calculating the determinant

results in $7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4) = -140 + 120 + 20 = 0$, verifying that the system has infinitely many solutions. Summing the first two equations, $7x + 6y - 2z = 0$ and $3x + 4y + 2z = 0$, leads to $10x + 10y = 0$, which simplifies to $y = -x$. Substituting this expression into the first equation, $7x + 6(-x) - 2z = 0$, yields $x - 2z = 0$, identifying that the infinite solution set must satisfy $x = 2z$.

The sum of the diagonal elements of the product $AA^T$ is equivalent to the sum of the squares of all entries in the matrix $A$, expressed as $\text{trace}(AA^T) = \sum a_{ij}^2 = 3$. Given that the entries are restricted to the set $\{-1, 0, 1\}$, squaring any element results in either $0$ or $1$, implying that exactly three positions in the $3 \times 3$ matrix must contain non-zero values, while the remaining six must be zero.

There are $\binom{9}{3}$ ways to choose the three positions for the non-zero entries from the nine available slots in the matrix. Each of these three chosen positions can independently hold a value of either $1$ or $-1$, providing $2^3$ different sign combinations for any given selection of positions. Combining these factors yields the final count of such matrices:
$\binom{9}{3} \times 2^3 = 672$

The solvability of a system of linear equations depends on the determinant $D$ of its coefficient matrix, which dictates whether the system possesses a unique solution or becomes degenerate. Evaluating the determinant for this system results in
$D = (\lambda+8)(2-\lambda)$
which reveals that the system's behavior changes at $\lambda = -8$ and $\lambda = 2$. Specifically, when $\lambda = 2$, the determinant $D$ becomes zero, requiring a check for consistency through the numerator determinant $D_1$; since $D_1 \neq 0$ at this value, the system is inconsistent and yields no solution.

The matrix $A$ represents a rotation operator, where powers follow the form

. Combining $A + A^4$ yields a matrix whose determinant simplifies via trigonometric identities to $\det(B) = 2 + 2\cos 3\theta$. Inserting $\theta = \frac{\pi}{5}$ into this expression results in $2 + 2\cos\frac{3\pi}{5}$. Because $\cos\frac{3\pi}{5}$ is equal to $-\sin 18^\circ$, which is $-\left(\frac{\sqrt{5}-1}{4}\right)$, the determinant computes to $\frac{5-\sqrt{5}}{2}$, or approximately $1.382$, which resides in the interval $(1, 2)$.

Simplifying complex determinants becomes much more manageable by performing row or column operations to introduce zeros, which makes the expansion process straightforward. In this specific case, observing the relationship between the first and second columns reveals that their difference is a constant vector. Applying the column operation $C_3 \rightarrow C_3 - (C_1 - C_2)$ transforms the third column into a vector containing only zeros in the first two positions and a constant value in the third. This operation results in a determinant that can be expanded easily along the third column:
$f(\theta) = \begin{vmatrix} -\sin^2\theta & -1-\sin^2\theta & 0 \\ -\cos^2\theta & -1-\cos^2\theta & 0 \\ 12 & 10 & -4 \end{vmatrix}$
By expanding this matrix, we obtain $-4$ multiplied by the determinant of the remaining $2 \times 2$ block. Calculating this block yields $(-\sin^2\theta)(-1-\cos^2\theta) - (-1-\sin^2\theta)(-\cos^2\theta)$, which simplifies to $\sin^2\theta - \cos^2\theta$. Multiplying this result by the $-4$ from the expansion gives $4(\cos^2\theta - \sin^2\theta)$, which is equivalent to the trigonometric identity $4\cos 2\theta$.

With the function simplified to $f(\theta) = 4\cos 2\theta$, we evaluate its range over the given domain $\theta \in [\frac{\pi}{4}, \frac{\pi}{2}]$. As $\theta$ varies within this interval, the argument $2\theta$ spans from $\frac{\pi}{2}$ to $\pi$. Within the interval $[\frac{\pi}{2}, \pi]$, the cosine function takes on all values between $0$ and $-1$. Multiplying these boundaries by $4$ confirms that the output of the function $f(\theta)$ lies between $-4$ and $0$. Therefore, with the minimum $m = -4$ and the maximum $M = 0$, the resulting ordered pair is $(-4, 0)$.

For a system of linear equations to be consistent when the determinant of the coefficient matrix is zero, the equations must satisfy a linear dependency where a weighted combination of the equations eliminates all variable terms. By inspecting the coefficients of $x$, $y$, and $z$, it becomes clear that doubling the first equation and adding it to the second and third equations cancels every variable, resulting in the identity $2(x - 4y + 7z) + (3y - 5z) + (-2x + 5y - 9z) = 0$. Because this specific linear combination reduces the variable side of the system to zero, the same weighted combination must be applied to the constant terms on the right-hand side, meaning $2g + h + k$ must equal zero for the system to remain consistent.

Since the determinant of an inverse matrix is the reciprocal of the determinant of the original, the condition $\det(A) + 1 = 0$ implies that the determinant of the given matrix must be $-1$. Expanding the determinant of the matrix yields $5(-2 - 3) - 2\alpha(0 - \alpha) + 1(0 - 2\alpha)$, which simplifies to $2\alpha^2 - 2\alpha - 25$. Setting this equal to $-1$ results in the quadratic equation $2\alpha^2 - 2\alpha - 24 = 0$, or $\alpha^2 - \alpha - 12 = 0$. Solving for $\alpha$ reveals the roots $4$ and $-3$, which sum to $1$.

When three numbers form an arithmetic progression, they share a constant common difference, allowing the determinant of the matrix to be expressed as a function of a single variable. Expanding the determinant of the matrix

while utilizing the property $2b = 2+c$ reveals that $|A| = 2(b-2)^3$. Applying the condition $|A| \in [2, 16]$ leads to the inequality $2 \le 2(b-2)^3 \le 16$, which simplifies to $1 \le (b-2)^3 \le 8$. Taking the cube root of this expression shows that $1 \le b-2 \le 2$, placing $b$ within the range $3 \le b \le 4$. Substituting these bounds into the relationship $c = 2b-2$ confirms that $c$ lies in the interval $[4, 6]$.