Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations is inconsistent when the determinant of the coefficient matrix equals zero, as this indicates that the equations either share no points of intersection or are contradictory. The determinant of the coefficient matrix for this system is defined by:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix} = 1(10\alpha-9\alpha) - 1(5\alpha-3) + 1(3\alpha^{2-}2\alpha) = 3\alpha^{2-}6\alpha+3$
Setting $\Delta=0$ yields $3(\alpha-1)^2=0$, which identifies $\alpha=1$ as the only potential value for inconsistency. Substituting $\alpha=1$ into the original system creates three equations where the first two imply $y+2z=-2$, while the last two imply $y+2z=5$; because these conditions are mutually exclusive, the system is inconsistent for exactly one value of $\alpha$.

The trace of $A^T A$ is equivalent to the sum of the squares of all entries in the matrix $A$, which, given the allowed entry set $\{-1, 0, 1\}$, simply counts the number of non-zero entries. For this sum to equal 6, the matrix must contain exactly 6 non-zero values and 3 zeros. Selecting the 3 positions for these zeros from the 9 total slots can be done in $\binom{9}{3}$ ways, while the 6 remaining positions can each be filled by either 1 or -1, providing $2^6$ distinct configurations. Multiplying these possibilities together, the total number of such matrices is $\binom{9}{3} \times 2^6 = 84 \times 64 = 5376$.

Squaring the matrix

reveals

, a relationship that allows all higher powers to be reduced. Since every even power $A^{2k}$ results in $(-4)^k I$ and every odd power $A^{2k-1}$ results in $(-4)^{k-1} A$, the sums $M$ and $N$ behave as scalar multiples of the identity matrix and matrix $A$, respectively. Specifically, $M$ becomes a scalar multiple of $I$ and $N$ becomes a scalar multiple of $A$.

When we calculate $N^2$, we obtain a scalar multiple of $A^2$, which in turn transforms back into a scalar multiple of the identity matrix $I$. Multiplying $M$ by $N^2$ consequently yields a result in the form $K I$, where $K$ is a constant. Because any matrix that is a scalar multiple of the identity matrix is symmetric and the specific constant $K$ is not equal to $\pm 1$, the product is a non-identity symmetric matrix.

Mastering how determinants interact with scalar multiplication and matrix products is essential for solving this problem. For an $n \times n$ matrix $M$ scaled by $k$, the determinant rule is $\det(kM) = k^n \det(M)$, and for any two matrices $X$ and $Y$, the property $\det(XY) = \det(X)\det(Y)$ holds. Since both matrices here are of order 2, we start by simplifying $\det(5(\det A)B)$ using $\det(A) = 2$ and $\det(B) = 3$. This simplifies to $\det(10B)$, which expands to $10^2 \det(B)$, yielding 300. We then substitute this into the full expression to get $\det(300 A^2)$. Applying the scaling rule again gives $300^2 \det(A^2)$. Because $\det(A^2)$ is $(\det A)^2$, this value is 4. Multiplying $300^2$ by 4 involves breaking 300 into prime factors, which leads to $3^2 \times 2^4 \times 5^4$. Finally, multiplying by 4, or $2^2$, results in $2^6 \times 3^2 \times 5^4$. Comparing this with the required form, we have:

$2^6 \cdot 3^2 \cdot 5^4 = 2^a \cdot 3^b \cdot 5^c$

This identification shows $a=6$, $b=2$, and $c=4$, so the final sum $a+b+c = 12$.

Understanding the relationship between a matrix determinant and its adjoint, specifically that $|\operatorname{adj}(M)| = |M|^{n-1}$ for an $n \times n$ matrix, is essential here, along with the scaling property $|kA| = k^n|A|$. Since $|A||B| = |I| = 1$ and $|A| = \frac{1}{8}$, it follows that $|B| = 8$.

The expression $|\operatorname{adj}(B \operatorname{adj}(2A))|$ simplifies to $|B \operatorname{adj}(2A)|^2$. By the multiplicative property of determinants, $|B \operatorname{adj}(2A)|$ expands to $|B| \cdot |\operatorname{adj}(2A)|$. Calculating the term $|2A|$ gives $|2A| = 2^3 |A| = 8 \cdot \frac{1}{8} = 1$. Therefore, $|\operatorname{adj}(2A)|$ becomes $|2A|^2 = 1^2 = 1$. Substituting these values leads to the calculation $(8 \cdot 1)^2$, which equals 64.

The classification of a matrix as symmetric or skew-symmetric depends on whether its transpose equals the original matrix or its negative, utilizing the fundamental rules $A^T=A$, $B^T=-B$, and $(XY)^T=Y^T X^T$. Calculating the transpose for $A^{4-}B^4$ gives $(A^T)^{4-}(B^T)^4 = A^{4-}(-B)^4 = A^{4-}B^4$, which is symmetric, while the transpose for $AB-BA$ yields $B^T A^T - A^T B^T = -BA - A(-B) = AB-BA$, confirming it is also symmetric.

Examining $AB+BA$ gives $(AB)^T+(BA)^T = B^T A^T + A^T B^T = -BA - AB = -(AB+BA)$, which is skew-symmetric. In contrast, the transpose of $B^{5-}A^5$ is $(B^T)^{5-}(A^T)^5 = (-B)^{5-}A^5 = -B^{5-}A^5$, which simplifies to $-(B^{5+}A^5)$ rather than $-(B^{5-}A^5)$, proving that this matrix is not skew-symmetric.

The matrix

has the property that its square is

. Because $n(n+1)$ is always an even integer, the condition $A^{n(n+1)} = I$ is equivalent to requiring $A^2 = I$ for the matrix to satisfy the relation for all $n$. For the squared matrix to be the identity, we must have $b^2 = 1$ and $-a + ab = 0$, which restricts $b$ to $1$ given that $b$ must be a positive integer from the set $\{1, 2, 3, \ldots, 100\}$. Since $b=1$ satisfies the equation $-a + ab = 0$ for every possible value of $a$, there are 100 such matrices in the intersection.

Calculating the determinant of a matrix raised to a power is simplified by the property $\det(A^{n}) = (\det A)^{n}$. Given

, the determinant of $A$ is found by calculating $(2)(-1) - (3)(0)$, which equals -2. Consequently, $\det(A^{4}) = (-2)^{4}$, which is 16.

Because $A$ is an upper triangular matrix, $A^{10}$ is also upper triangular with diagonal elements $2^{10}$ and $(-1)^{10} = 1$. The matrix $2A$ is

, and its adjoint $\operatorname{Adj}(2A)$ is

, which is also upper triangular. Raising an upper triangular matrix to a power results in another upper triangular matrix, meaning $[\operatorname{Adj}(2A)]^{10}$ is upper triangular with diagonal elements $(-2)^{10} = 2^{10}$ and $4^{10} = 2^{20}$. The difference $A^{10} - [\operatorname{Adj}(2A)]^{10}$ is therefore an upper triangular matrix where the first entry on the main diagonal is $2^{10} - 2^{10} = 0$. Since the determinant of any upper triangular matrix is the product of its diagonal elements, the determinant of this difference is 0. Adding the two parts together yields $16 + 0$, which is 16.

The inverse of a square matrix can be expressed as a linear combination of the identity matrix and the matrix itself, which allows us to determine the coefficients $\alpha$ and $\beta$ by equating individual matrix entries. Starting with the determinant $|A| = (1)(4) - (-1)(2) = 6$, we find the inverse matrix using the adjugate method, yielding

. We define the linear combination $\alpha I + \beta A$ as

, which simplifies to the single matrix

. Equating the elements of this matrix to the inverse $A^{-1}$, we obtain $2\beta = -\frac{1}{3}$, meaning $\beta = -\frac{1}{6}$, and $\alpha + \beta = \frac{2}{3}$, implying $\alpha - \frac{1}{6} = \frac{2}{3}$ or $\alpha = \frac{5}{6}$. Substituting these results into the final expression, $4(\alpha - \beta) = 4(\frac{5}{6} - (-\frac{1}{6})) = 4(1) = 4$.

The trace of the matrix product $M^T M$ is equivalent to the sum of the squares of all entries within $M$. For a $3 \times 3$ matrix, which contains 9 total entries, the problem specifies that this sum of squares must be 7, with each entry restricted to the set $\{0, 1, 2\}$.

Representing the count of entries equal to 0, 1, and 2 as $n_0$, $n_1$, and $n_2$, respectively, these counts must satisfy the conditions $n_0 + n_1 + n_2 = 9$ and $n_1 + 4n_2 = 7$.

If we set $n_2 = 0$, then $n_1 = 7$, which implies $n_0 = 2$. The number of unique arrangements is given by the expression $\frac{9!}{0! 7! 2!} = 36$. If we set $n_2 = 1$, then $n_1 = 7 - 4 = 3$, which leaves $n_0 = 9 - 1 - 3 = 5$. The number of unique arrangements is given by the expression $\frac{9!}{1! 3! 5!} = 504$.

Setting $n_2$ to any value greater than or equal to 2 would cause the sum of squares to exceed 7, meaning no other cases exist. Combining the 36 arrangements from the first case and the 504 arrangements from the second case, there are 540 such matrices in total.

The determinant of a matrix

is calculated as $ad - bc$. Since each entry is restricted to the set $\{\pm 3, \pm 2, \pm 1, 0\}$, the maximum magnitude for any product of two entries is $3 \times 3 = 9$. To satisfy the condition $ad - bc = 15$, we must identify pairs of products $ad$ and $bc$ that yield a difference of 15 while remaining within the bounds of possible product values.

One scenario requires $ad = 9$ and $bc = -6$. The possible pairs for $(a, d)$ such that $ad = 9$ are $(3, 3)$ and $(-3, -3)$, providing 2 combinations. The possible pairs for $(b, c)$ such that $bc = -6$ are $(3, -2), (-3, 2), (-2, 3),$ and $(2, -3)$, providing 4 combinations. Multiplying these possibilities yields $2 \times 4 = 8$ matrices.

A second scenario occurs when $ad = 6$ and $bc = -9$. For $ad = 6$, the valid pairs are $(3, 2), (2, 3), (-3, -2),$ and $(-2, -3)$, resulting in 4 combinations. For $bc = -9$, the valid pairs are $(3, -3)$ and $(-3, 3)$, resulting in 2 combinations. This combination yields another $4 \times 2 = 8$ matrices. Adding these two groups gives a total of 16 matrices.

Three terms form an arithmetic progression when the middle term is the arithmetic mean of the flanking terms, which means doubling the middle term equals the sum of the others. Applying this, the condition $2 \log_{10}(4^x - 2) = 1 + \log_{10}(4^x + \frac{18}{5})$ simplifies using logarithm properties to $\log_{10}(4^x - 2)^2 = \log_{10}(10 \cdot 4^x + 36)$ after noting that $1 = \log_{10} 10$.

Equating the arguments gives $(4^x - 2)^2 = 10 \cdot 4^x + 36$, which expands to $(4^x)^2 - 4 \cdot 4^x + 4 = 10 \cdot 4^x + 36$. Rearranging leads to the quadratic equation $(4^x)^2 - 14 \cdot 4^x - 32 = 0$, which factors as $(4^x - 16)(4^x + 2) = 0$. Since $4^x$ must be positive for any real $x$, we discard the negative root and identify $4^x = 16$, which confirms $x = 2$.

Substituting $x = 2$ into the provided matrix yields the determinant

. Expanding along the first row provides $3(0 \cdot 0 - 2 \cdot 1) - 1(1 \cdot 0 - 2 \cdot 2) + 4(1 \cdot 1 - 0 \cdot 2)$, which simplifies to $-6 + 4 + 4$, resulting in a final value of 2.

When raising a matrix to a power, calculating the initial few terms is an effective method to establish a recurring pattern. By computing

and

, we see that the diagonal elements remain $1$ and the top-right element remains $0$, while the bottom-left element follows the form $\frac{n}{2}$ for any power $n$. Extending this rule to $P^{50}$, the bottom-left entry becomes $\frac{50}{2}$, which is $25$. This leads directly to the resulting matrix

.

The area of a triangle with vertices at $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is determined by half the absolute value of the determinant of the matrix formed by the coordinates and a column of ones. Applying this principle to the points $A(a, 0)$, $B(b, 2b+1)$, and $C(0, b)$ involves evaluating the expression $\frac{1}{2} |a(b+1) + b^2| = 1$.

This equality implies two distinct scenarios because the internal determinant value can be either positive or negative to result in an absolute value of 1. Specifically, the expression $a(b+1) + b^2$ must equal either $2$ or $-2$. Solving the first case for $a$ yields $a = \frac{2-b^2}{b+1}$, while solving the second case yields $a = \frac{-2-b^2}{b+1}$.

To find the sum of all possible values for $a$, these two expressions are added together: $\frac{2-b^2}{b+1} + \frac{-2-b^2}{b+1}$. Combining the numerators results in $2 - b^2 - 2 - b^2$, which simplifies to $-2b^2$. Consequently, the total sum is $\frac{-2b^2}{b+1}$.

When evaluating a system of linear equations to determine its solution type, the determinant of the coefficient matrix serves as the primary indicator. Calculating the determinant for this specific system, we find that the result is $|A| = 0$, which signifies that the system cannot possess a unique solution and must instead have either no solution or infinitely many solutions.

To distinguish between these two possibilities, we examine the auxiliary determinants $|A_1|$, $|A_2|$, and $|A_3|$ formed by replacing the columns of the coefficient matrix with the constants $a$, $b$, and $c$. Computation of these determinants shows that $|A_1| = 4(5a - 2b - c)$, $|A_2| = -5(5a - 2b - c)$, and $|A_3| = -2(5a - 2b - c)$. Because these auxiliary determinants depend entirely on the expression $5a - 2b - c$, setting this expression to zero ensures that all determinants in the system vanish, which is the necessary condition for the system to have an infinite number of solutions.

The sequence $a_r = e^{i 2\pi r/9}$ forms a geometric progression where each term can be expressed as $a_r = \lambda^r$ using the common ratio $\lambda = e^{i 2\pi/9}$. This exponential structure allows for the simplification of products by summing their exponents, so the expression $a_1 a_9 - a_3 a_7$ becomes $\lambda^1 \cdot \lambda^9 - \lambda^3 \cdot \lambda^7$, which evaluates to $\lambda^{10} - \lambda^{10} = 0$. Since the determinant of a matrix whose entries are terms from a geometric progression evaluates to zero, identifying the option that simplifies to this null value points directly to the correct relationship.

The fundamental requirement for the matrix $A$ in this equation is derived from expanding the expression $(I-A)^3 = I-A^3$. The expansion on the left yields
$I - 3A + 3A^2 - A^3$
and setting this equal to $I - A^3$ simplifies the problem significantly to the condition $3A^2 - 3A = 0$, which implies $A^2 = A$. This means the matrix $A$ must be idempotent.

Considering the matrix

, the condition $A^2 = A$ requires that
$\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} = \begin{pmatrix} a^2 & ab+bd \\ 0 & d^2 \end{pmatrix} = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$
This equality leads to three distinct conditions: $a^2=a$, $d^2=d$, and $b(a+d)=b$. The equations $a^2=a$ and $d^2=d$ restrict the possible values for $a$ and $d$ to 0 or 1, as the square of these values is equal to the value itself.

We can now branch the analysis based on the value of $b$. If $b=0$, the third condition $b(a+d)=b$ is satisfied for any choice of $a$ and $d$ where both are 0 or 1. There are four such pairs $(a,d)$: $(0,0), (0,1), (1,0),$ and $(1,1)$, giving us four distinct matrices.

Alternatively, if $b$ is non-zero, it must be either 1 or -1 based on the provided set, and the equation $b(a+d)=b$ simplifies to $a+d=1$. The pairs $(a,d)$ that satisfy $a+d=1$ are $(1,0)$ and $(0,1)$. Since there are two possible values for $b$ for each of these two pairs, this provides another four solutions, resulting in 8 total elements.

For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must equal zero. Expanding the determinant of the matrix formed by the root coefficients yields $3\alpha\beta\gamma - (\alpha^3 + \beta^3 + \gamma^3) = 0$, which factors into the expression $-(\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) = 0$.

Given the condition $a \neq 0$, the sum of the roots $\alpha + \beta + \gamma = -a$ cannot be zero, which forces the term $\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha$ to equal zero. This equality is satisfied if and only if $\alpha = \beta = \gamma$.

With all roots being equal, the relations between the coefficients and the roots are defined as $-a = 3\alpha$ and $b = 3\alpha^2$. Substituting these expressions into $\frac{a^2}{b}$ yields $\frac{(-3\alpha)^2}{3\alpha^2} = \frac{9\alpha^2}{3\alpha^2} = 3$.

The matrix $A$ is orthogonal, meaning that $AA^T = I$, which significantly simplifies the expression for $Q^{2021}$. Given the definition $Q = A^T B A$, the product $Q^2$ expands to $(A^T B A)(A^T B A)$, which simplifies to $A^T B (AA^T) B A = A^T B^2 A$. By induction, this pattern continues so that $Q^{2021} = A^T B^{2021} A$. Consequently, substituting this into the required expression $A Q^{2021} A^T$ yields $A(A^T B^{2021} A)A^T$, which reduces simply to $B^{2021}$ because the product $AA^T$ is the identity matrix.

For the matrix

, raising it to the power of 2021 follows an additive pattern for the off-diagonal element, resulting in

. Calculating the inverse of this matrix requires dividing the adjoint by the determinant, which is 1, simply requiring the negation of the off-diagonal term. Therefore, the inverse of the matrix $A Q^{2021} A^T$ is
$\begin{pmatrix} 1 & 0 \\ -2021i & 1 \end{pmatrix}$

The sum of the diagonal entries of the product matrix $A A^T$ is identical to the sum of the squares of all entries within the $3 \times 3$ matrix $A$. Since each of the nine entries must be chosen from the set $\{0, 1, 2, 3\}$, the problem requires identifying all combinations of nine values whose squares add up to exactly 9.

The simplest case is when every entry in the matrix is 1, yielding a sum of $1^2 \times 9 = 9$, which can occur in only 1 way. Another scenario occurs when one entry is 3 and the remaining eight entries are 0, resulting in a sum of $3^2 + 8(0^2) = 9$; there are $\frac{9!}{8!} = 9$ such matrices.

We can also satisfy the sum by selecting one 1, two 2s, and six 0s, as the squares combine to $1^2 + 2(2^2) + 6(0^2) = 9$. The number of distinct arrangements for this case is $\frac{9!}{6! 2! 1!} = 252$. Finally, considering the case with one 2, five 1s, and three 0s, the squares total $2^2 + 5(1^2) + 3(0^2) = 9$. This configuration results in $\frac{9!}{5! 3! 1!} = 504$ distinct matrices. Summing these possibilities gives a total of $1 + 9 + 252 + 504 = 766$.