Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The equation $x^{2+}x+1=0$ indicates that $\alpha$ is a primitive complex cube root of unity, which implies the properties $\alpha^3=1$ and $\alpha^{2+}\alpha+1=0$. Performing the matrix multiplication

generates the linear system $4-a-2b=0$ and $64-a-14b=0$. Subtracting the first equation from the second yields $12b=60$, which gives $b=5$, and substituting this back into the first equation results in $a=-6$.

Replacing these values in the provided expression $\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3$ yields $\frac{4}{\alpha^4} + m\alpha^6 + \frac{n}{\alpha^5} = 3$. Utilizing the identities $\alpha^3=1$, $\frac{1}{\alpha}=\alpha^2$, and $\frac{1}{\alpha^2}=\alpha$ reduces this to $4\alpha^2 + m + n\alpha = 3$, or $4\alpha^2 + n\alpha + (m-3) = 0$. Since this polynomial in $\alpha$ must vanish for the same roots as $\alpha^{2+}\alpha+1=0$, it follows that $4\alpha^2 + n\alpha + (m-3)$ is a scalar multiple of $\alpha^{2+}\alpha+1$. Comparing coefficients gives $n=4$ and $m-3=4$, leading to $m=7$ and a sum of $m+n=11$.

A system of linear equations possesses infinitely many solutions when the determinant of the coefficient matrix and the determinants of the augmented matrices are all equal to zero. This condition ensures that the system is consistent and that the equations are linearly dependent.

By setting the determinant $\Delta_2$ to zero, we determine the value of $\mu$:
$\Delta_2 = \begin{vmatrix} 2 & 3 & 5 \\ 3 & -1 & 7 \\ 4 & \mu & 9 \end{vmatrix} = 0$
Evaluating this determinant leads to the linear equation $\mu + 5 = 0$, which yields $\mu = -5$. Similarly, computing the determinant $\Delta_3$ and setting it to zero allows us to find the value of $\lambda$:
$\Delta_3 = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0$
This calculation results in the equation $\lambda + 1 = 0$, identifying $\lambda = -1$.

With the values of $\lambda$ and $\mu$ established, calculating $\lambda^{2+}\mu^2$ involves evaluating the expression:
$\lambda^{2+}\mu^2 = (-1)^2 + (-5)^2$
Substituting the squared values provides 1 + 25, which results in a final value of 26.

Variance quantifies the dispersion of a random variable $X$ relative to its expected value, defined by the formula $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$. For 16 total possible matrices formed by entries of 0 and 1, the determinant $X$ takes the value 0 in 10 instances, 1 in 3 instances, and -1 in 3 instances. Calculating the mean yields $E(X) = 0 \cdot \frac{10}{16} + 1 \cdot \frac{3}{16} + (-1) \cdot \frac{3}{16} = 0$. Because the mean is zero, the variance simplifies to the second moment $E(X^2) = 0^2 \cdot \frac{10}{16} + 1^2 \cdot \frac{3}{16} + (-1)^2 \cdot \frac{3}{16}$, which results in $\frac{3}{16} + \frac{3}{16} = \frac{6}{16}$, or $\frac{3}{8}$.

The most effective way to evaluate a complex determinant is to simplify it using row and column transformations that create a sparse matrix, which then allows for straightforward expansion. By applying the operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$, the determinant becomes

. Performing the further operation $C_2 \rightarrow C_1 + C_2$ leads to the matrix

, which simplifies to $f(x) = 2 + 4\sin 4x$. Since $\sin 4x$ varies between $-1$ and $1$, the term $4\sin 4x$ fluctuates between $-4$ and $4$, establishing the maximum value $M = 6$ and the minimum value $m = -2$. Subtracting the fourth power of the minimum from the fourth power of the maximum results in $6^4 - (-2)^4 = 1296 - 16$, which equals $1280$.

To find the inverse of the matrix product $A \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right)^{-1} B$, apply the reversal law $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$, which transforms the expression into $B^{-1} \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right) A^{-1}$. Since the adjoint of the inverse of a non-singular matrix $M$ is defined as $\operatorname{adj}(M^{-1}) = \frac{M}{|M|}$, the terms inside the parentheses become $\frac{A}{|A|} + \frac{B}{|B|}$. Distributing $B^{-1}$ from the left and $A^{-1}$ from the right results in $\frac{B^{-1} A A^{-1}}{|A|} + \frac{B^{-1} B A^{-1}}{|B|}$, which simplifies to $\frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$ because $A A^{-1}$ and $B^{-1} B$ represent the identity matrix. Replacing the inverses $B^{-1}$ and $A^{-1}$ with their adjoint forms $\frac{\operatorname{adj}(B)}{|B|}$ and $\frac{\operatorname{adj}(A)}{|A|}$ yields $\frac{\operatorname{adj}(B)}{|B||A|} + \frac{\operatorname{adj}(A)}{|A||B|}$, which results in the final expression $\frac{1}{|AB|} ( \operatorname{adj}(B) + \operatorname{adj}(A) )$.

A singular matrix of order $2 \times 2$ must satisfy the condition $ad = bc$. Categorizing by the number of distinct elements used from the set {2, 3, 6, 9} reveals the total count. When only one distinct number is used, all four positions in the matrix must be identical, leading to 4 possible matrices. Selecting exactly two distinct numbers allows for 6 combinations, and each combination produces 4 valid arrangements, resulting in 24 total matrices.

It is impossible to satisfy the condition using exactly three distinct numbers, so this case contributes 0 matrices. When all four distinct numbers are used, the determinant condition is satisfied by the relations $2 \times 9 = 3 \times 6 = 18$, which yields 8 valid arrangements. Summing these results gives $4 + 24 + 0 + 8 = 36$ singular matrices.

The property that for any $n \times n$ matrix $M$, the double adjoint is expressed by the following identity:
$\operatorname{adj}(\operatorname{adj}(M)) = |M|^{n-2} M$
Applying this identity to the $3 \times 3$ matrix $2A$ with $n=3$, we obtain $B = |2A|^{3-2} (2A)$, which simplifies to $B = |2A| \cdot 2A$. Since scaling a $3 \times 3$ matrix by a factor of $2$ increases its determinant by a factor of $2^3$, we find that $|2A| = 8|A| = 8 \times \frac{1}{2} = 4$. Substituting this value into the expression for $B$ yields $4 \times 2A = 8A$.

With $B$ established as $8A$, finding the required values becomes a straightforward application of matrix properties. The trace operator follows linearity, such that $\operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \operatorname{trace}(A) = 8 \times 3 = 24$. Similarly, the determinant of a $3 \times 3$ matrix scaled by $8$ is given by $|B| = |8A| = 8^3 |A|$, which calculates to $512 \times \frac{1}{2} = 256$. Adding these results produces the final value of $24 + 256 = 280$.

The fundamental principle here is that $P$ is an orthogonal matrix, satisfying the condition $P^\top P = I$. When we define $B = P A P^\top$, raising $B$ to any power $k$ results in a telescopic cancellation of the inner matrices, yielding $B^k = P A^k P^\top$. Therefore, the expression for $C$ simplifies as follows:
$C = P^\top (P A^{10} P^\top) P = (P^\top P) A^{10} (P^\top P) = I A^{10} I = A^{10}$

Since $A$ is an upper triangular matrix, its $10^{th}$ power is also upper triangular. The diagonal elements of $A^{10}$ are simply the diagonal entries of $A$ raised to the power of 10, which are $(\frac{1}{\sqrt{2}})^{10}$ and $1^{10}$. Calculating these, we get $\frac{1}{32}$ and $1$. Adding these diagonal elements gives $\frac{1}{32} + 1 = \frac{33}{32}$, which corresponds to $\frac{m}{n}$ where $\gcd(33, 32) = 1$. The sum $m + n = 33 + 32 = 65$.

Determinant properties for a $3 \times 3$ matrix $A$ dictate that $\det(k A) = k^3 \det(A)$ and $\det(\operatorname{adj}(A)) = (\det(A))^2$. Applying these to the expression $\det(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))$ requires evaluating the determinant from the outside inward. The outermost factor of 3 scales the determinant by $3^3$, while the adjoint operation squares the determinant of its argument, simplifying the expression to $3^3 \cdot ( \det(-6 \operatorname{adj}(3 A)) )^2$. Within the next layer, the factor $-6$ scales the determinant by $(-6)^3$, while $\det(\operatorname{adj}(3 A))$ becomes $(\det(3 A))^2$.

This results in the expression $3^3 \cdot ( (-6)^3 \cdot (\det(3 A))^2 )^2$, which simplifies to $3^3 \cdot (-6)^6 \cdot (\det(3 A))^4$. Given $\det(3 A) = 3^3 \det(A) = 27(-2) = -54$, the numerical calculation proceeds as $3^3 \cdot (2 \cdot 3)^6 \cdot (-54)^4$. Expanding the powers further, $(-54)^4 = (2 \cdot 3^3)^4 = 2^4 \cdot 3^{12}$, which makes the full product $3^3 \cdot 2^6 \cdot 3^6 \cdot 2^4 \cdot 3^{12} = 2^{10} \cdot 3^{21}$. Comparing this result to the given form $2^{m+n} \cdot 3^{mn}$ establishes the system of equations $m+n=10$ and $mn=21$. With the condition $m > n$, the only solution is $m=7$ and $n=3$, yielding $4m + 2n = 4(7) + 2(3) = 34$.

Counting the matrices within these sets relies on identifying independent entry choices under specific structural constraints. For symmetric matrices in $S_1$, defined by $A = A^T$, only six positions are independent-the three diagonal elements and three above-diagonal elements-while the remaining entries are fixed by reflection. Since each of these six positions can be filled by any of the five values in $S$, there are $5^6$ such matrices. Skew-symmetric matrices in $S_2$, defined by $A = -A^T$, require every diagonal element to be zero. Because zero is not contained within $S$, no such matrices exist, resulting in a count of zero for $S_2$.

The set $S_3$ consists of matrices with a trace of zero, meaning the diagonal sum $a_{11} + a_{22} + a_{33}$ must equal zero. The six off-diagonal entries can be any of the five values, providing $5^6$ total combinations for those positions. For the diagonal entries, we identify triples that sum to zero using values from $S$: the set $(1, 2, -3)$ has $3! = 6$ permutations, $(1, 1, -2)$ has $3!/2! = 3$ permutations, and $(-1, -1, 2)$ has $3!/2! = 3$ permutations, totaling $12$ valid diagonal combinations. Thus, the total count for $S_3$ is $12 \times 5^6$.

To compute the size of the union $n(S_1 \cup S_2 \cup S_3)$, we apply the Principle of Inclusion-Exclusion. Given $S_2$ is empty, all intersections involving it are also empty, simplifying the calculation to $n(S_1) + n(S_3) - n(S_1 \cap S_3)$. The intersection $S_1 \cap S_3$ represents symmetric matrices with a trace of zero. In this case, the off-diagonal entries provide only three independent choices, $5^3$, while the diagonal still allows for the $12$ valid combinations found earlier, yielding $12 \times 5^3$ matrices. Combining these, the total is $5^6 + 12 \times 5^6 - 12 \times 5^3$, which simplifies to $13 \times 5^6 - 12 \times 5^3$. Factoring out $125$ as $5^3$ gives $125(13 \times 5^3 - 12)$, which equals $125(1625 - 12)$, confirming that $\alpha$ is $1613$.

For a system of linear equations to possess infinitely many solutions, the determinant of the coefficient matrix must vanish, as this signifies that the equations are linearly dependent and do not represent a unique intersection point. The determinant must satisfy the following:

$\begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 12 & 3 & -(4+\lambda) \end{vmatrix} = 0$

Solving this expression yields $15\lambda - 75 = 0$, confirming $\lambda = 5$.

Substituting $\lambda=5$ into the system, the condition for consistency is met by ensuring the augmented matrix determinant also vanishes, which leads to $21\mu - 189 = 0$, identifying $\mu = 9$. The radius of the circle centered at $(5, 9)$ and touching the line $4x - 3y = 0$ is the perpendicular distance from the point to the line, calculated as:

$r = \frac{|4(5) - 3(9)|}{\sqrt{4^2 + 3^2}} = \frac{7}{5}$

For a system of linear equations to possess infinitely many solutions, the determinant of the coefficient matrix $D$ must equal zero, as must the determinants $D_1$, $D_2$, and $D_3$ obtained by replacing the columns of the coefficient matrix with the constant terms on the right side of the equations. Evaluating the coefficient determinant

yields the expression $2a+b-6$, which must satisfy $2a+b=6$ for the system to be consistent. Similarly, calculating the determinant

results in the expression $2a+2b-16$, which implies the condition $a+b=8$.

Solving this system of two linear equations, $2a+b=6$ and $a+b=8$, identifies the values $a=-2$ and $b=10$. Substituting these values into the expression $7a+3b$ results in $7(-2) + 3(10)$, which simplifies to a final value of $16$.

Evaluating the limit of the function as $x$ approaches zero relies on the fundamental trigonometric identity $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Substituting this value into the determinant gives the expression
$\begin{vmatrix} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{vmatrix}$
Applying row transformations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ simplifies the matrix to
$\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ a & 1 & b+1 \end{vmatrix}$
Expanding this determinant yields $a+b+2$, which allows us to equate terms to find $\lambda=2$, $\mu=1$, and $v=1$. Consequently, the square of the sum $(\lambda+\mu+v)^2 = (2+1+1)^2$ results in 16.

A system of linear equations is inconsistent if the reduction process leads to a contradiction, where the variables are eliminated but the remaining constants do not equate to zero. Starting with the given system, we simplify the rows by subtracting the first equation from the second to find $y+4z=3$ and from the third to obtain $4y+(\lambda-1)z = \mu-6$. By isolating the variables in the final step, we perform the operation of subtracting 4 times the first derived equation from the second derived equation, which yields $(\lambda-17)z = \mu-18$. This statement results in a mathematical contradiction, indicating no solution, precisely when the coefficient of $z$ is zero and the right side is non-zero, requiring $\lambda=17$ and $\mu\neq 18$.

A system of linear equations admits infinitely many solutions if the determinant of the coefficient matrix is zero, and the determinants formed by substituting the constants into the columns of the coefficient matrix are also zero. This condition ensures that the system is dependent and that the geometric planes defined by these equations intersect along a common line rather than at a single point or not at all.

Calculating the determinant of the coefficient matrix

leads to the equation $\lambda\mu+42\lambda-4\mu+107=0$. To find the specific values of the parameters, examine the determinant

, which involves substituting the constants into the second column. Expanding this determinant yields $1(5\mu-165)-2(2\mu-70)-3(66-70)=0$, which simplifies to $\mu-13=0$, confirming that $\mu=13$. Substituting $\mu=13$ back into the original determinant equation results in $13\lambda+42\lambda-4(13)+107=0$. Simplifying this expression gives $55\lambda+55=0$, which reveals that $\lambda=-1$. The sum of these two variables is therefore $-1+13$, which equals 12.

When a system of linear equations possesses infinitely many solutions, the determinants used in Cramer's rule-specifically the numerator determinants $\Delta x$, $\Delta y$, and $\Delta z$-must all vanish. This condition signifies that the system is consistent and dependent, allowing for the determination of the unknown variables $\lambda$ and $\mu$.

Setting the determinant

to zero yields $2(212\lambda + 564) + 1(1060 - 1200) + 4(-235 - 100\lambda) = 0$. Simplifying this expression results in $424\lambda + 1128 - 140 - 940 - 400\lambda = 0$, which reduces to $24\lambda + 48 = 0$, confirming that $\lambda = -2$. With $\lambda$ established, evaluate the determinant

, which expands to $2(-2\mu + 141) + 1(5\mu - 300) + 1(-235 + 200) = 0$. Further simplification gives $-4\mu + 282 + 5\mu - 300 - 35 = 0$, or $\mu - 53 = 0$, meaning $\mu = 53$. Substituting these values into the required expression $\mu - 2\lambda$ yields $53 - 2(-2)$, which equals 57.

The expression $C_{ij} = \sum_{k=1}^{2} a_{ik} A_{jk}$ represents the product of matrix $A$ and the transpose of its cofactor matrix, which is known as the adjoint of $A$, or $\text{adj}(A)$. A fundamental property of square matrices states that the product of any matrix and its adjoint equals the determinant of that matrix multiplied by the identity matrix, expressed as $A \cdot \text{adj}(A) = |A|I$. Consequently, $C$ must be a diagonal matrix where each diagonal element is the determinant $|A|$.

Simplifying the entries of matrix $A$ yields $a_{11} = \log_5 128 = 7 \log_5 2$, $a_{12} = \log_4 5 = \frac{1}{2} \log_2 5$, $a_{21} = \log_5 8 = 3 \log_5 2$, and $a_{22} = \log_4 25 = \log_2 5$. Calculating the determinant results in $|A| = (7 \log_5 2)(\log_2 5) - (\frac{1}{2} \log_2 5)(3 \log_5 2)$. Since $\log_5 2 \cdot \log_2 5 = 1$, the expression simplifies to $7 - \frac{3}{2} = \frac{11}{2}$. Given that $C$ is a $2 \times 2$ scalar matrix with $\frac{11}{2}$ on its diagonal, its determinant is $|C| = (\frac{11}{2})^2 = \frac{121}{4}$. The requested value $8 |C|$ is $8 \cdot \frac{121}{4} = 242$.

The calculation of determinants relies on fundamental properties, specifically that $\det(kM) = k^n \det(M)$ for an $n \times n$ matrix and $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$. Isolating matrix $A$ from the equation

results in

. Finding the determinant of this matrix as $-2a + 2$ and equating it to the provided value of $-4$ gives $-2a + 2 = -4$, which simplifies to $a = 3$.

Substituting $a = 3$ into the expression $\det((a+1) \operatorname{adj}((a-1) A))$ yields $\det(4 \operatorname{adj}(2A))$. By applying the determinant properties mentioned, this expression expands to $4^3 (\det(2A))^2$. Since $\det(2A) = 2^3 \det(A) = 8 \times (-4) = -32$, substituting this value leads to $4^3 \times (-32)^2$, which is $64 \times 1024 = 65536$. Representing $65536$ as $2^{16} \times 3^0$ confirms that $m = 16$ and $n = 0$, leading to a total value of $m + n = 16$.

The value of the determinant of the matrix $A$ is $8$, a result derived by simplifying the rows and columns to eliminate the variables $p$ and $q$. For any $3 \times 3$ matrix $M$, the determinant of its double adjoint follows the property $|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(3-1)^2} = |M|^4$. Because the matrix we are evaluating is $3A$, we first apply the scalar property of determinants, $|kA| = k^3|A|$, to find $|3A| = 3^3 \cdot |A| = 27 \cdot 8$, which simplifies to $216$ or $6^3$.

Substituting this determinant into our adjoint formula yields $|3A|^4 = (6^3)^4 = 6^{12}$. Expressing this value in terms of its prime factors, we obtain $2^{12} \cdot 3^{12}$. Comparing this result to the required form $2^m \cdot 3^n$, we can determine that $m=12$ and $n=12$. Adding these two integers gives $m+n=24$.

The property that lines $ax+by+c=0$ are concurrent for any three distinct consecutive terms of an arithmetic progression relies on the relation $2b=a+c$. By substituting $c = 2b-a$ into the line equation, we obtain $ax+by+2b-a=0$, which can be rearranged as $a(x-1) + b(y+2) = 0$. Since this equation must hold for any values of $a$ and $b$, the coefficients must independently satisfy $x-1=0$ and $y+2=0$, identifying point $P$ as $(1, -2)$.