Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The matrix relationship $P Q = k I_3$ implies that $Q = k P^{-1}$, where $P^{-1} = \frac{1}{|P|} \text{adj}(P)$. Calculating the determinant of $P$ by expanding along the first row gives $|P| = 3(0 - (-5\alpha)) - (-1)(2(0) - \alpha(3)) + (-2)(2(-5) - 0(3))$, which simplifies to $15\alpha - 3\alpha + 20 = 12\alpha + 20$.

The entry $q_{23}$ of the matrix $Q$ is defined by $\frac{k}{|P|} C_{32}$, where $C_{32}$ is the cofactor of the element in the third row and second column of $P$. Computing this cofactor yields $C_{32} = -((3)(\alpha) - (-2)(2)) = -(3\alpha + 4) = -3\alpha - 4$. Setting this expression equal to the given value $-\frac{k}{8}$ results in the equation $\frac{k(-3\alpha - 4)}{12\alpha + 20} = -\frac{k}{8}$. Since $k$ is non-zero, this simplifies to $\frac{3\alpha + 4}{12\alpha + 20} = \frac{1}{8}$, which leads to $24\alpha + 32 = 12\alpha + 20$, confirming that $\alpha = -1$.

Substituting $\alpha = -1$ into the determinant expression yields $|P| = 12(-1) + 20 = 8$. Using the determinant property $|Q| = |k P^{-1}| = k^3 |P|^{-1}$ and the given value $|Q| = \frac{k^2}{2}$, we have the equation $\frac{k^3}{8} = \frac{k^2}{2}$. Dividing by $k^2$ shows that $\frac{k}{8} = \frac{1}{2}$, resulting in $k = 4$. The final value is $\alpha^2 + k^2 = (-1)^2 + 4^2 = 17$.

For any square system of linear equations represented as $AX=B$, the existence of a unique solution is guaranteed if the determinant of the coefficient matrix $A$ is non-zero. By extracting the coefficients from the given equations, we define

. Computing the determinant results in $|A| = 2(8 - (-2)) - 3(12 - 2) + 2(-3 - 2)$, which simplifies to $20 - 30 - 10 = -20$. Since the determinant is $-20$, which is not equal to zero, the system must possess a unique solution.

Finding powers of the matrix $A$ reveals a consistent arithmetic progression in the first column of the second row. By calculating the first few powers, we see that

,

,

, and

. This pattern establishes the general form

.

Using this formula, we determine

and

. Subtracting these two matrices yields

, which simplifies to

. Evaluating the first option, $A^6 - A$, results in

, which produces the same resulting matrix

.

Decomposing the matrix $A$ into the sum of the identity matrix $I$ and a nilpotent matrix $C$ provides a path to use the binomial theorem for expanding powers of $A$. By writing $A = I + C$ with
$C = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$
it is evident that

and all higher powers $C^k$ for $k \geq 3$ result in the zero matrix. Since $I$ and $C$ commute, the binomial expansion $A^n = I + nC + \binom{n}{2}C^2$ holds perfectly.

Applying this expansion to $B = 7A^{20} - 20A^{7} + 2I$, the value of the entry $b_{13}$ is extracted by focusing on the $(1,3)$ position of each term. Both the identity matrix $I$ and the matrix $C$ contribute zero to this specific position, while the term $\binom{n}{2}C^2$ contributes $\binom{n}{2}$ because the $(1,3)$ entry of $C^2$ is $1$. Thus, the expression for $b_{13}$ simplifies to $7 \binom{20}{2} - 20 \binom{7}{2}$. Calculating these combinations as $7(190) - 20(21)$ yields $1330 - 420$, which equals 910.

Systems of linear equations containing a variable parameter fail to possess a unique solution precisely when the determinant of their coefficient matrix is zero. For the provided set of equations, the coefficient matrix is:

$\begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$

Computing this determinant expands to the cubic polynomial $k^3 - 3k + 2$. Setting this expression equal to zero and factoring yields $(k-1)^2(k+2) = 0$, indicating that the critical values to investigate are $k = 1$ and $k = -2$. When $k = 1$, the equations simplify into the identical plane $x + y + z = 1$, resulting in infinitely many solutions. Testing $k = -2$ introduces an inconsistency across the equations, confirming that this value yields no solution.

A symmetric $2 \times 2$ matrix with integer entries $a, b, c$ is represented as

, and calculating its square shows that the sum of the diagonal elements is $a^2 + b^2 + b^2 + c^2$. Setting this trace equal to 1 provides the equation $a^2 + 2b^2 + c^2 = 1$. Because $a, b, c$ are integers, their squares must be non-negative; therefore, $b$ must be 0 to prevent the sum from exceeding 1, which simplifies the condition to $a^2 + c^2 = 1$. This leaves only the integer solutions where one variable is $\pm 1$ and the other is 0, resulting in exactly four possible matrices:

,

,

, and

.

The trace operator is linear, which allows one to determine the trace of individual matrices by applying the sum of diagonal elements directly to the given equations rather than calculating the elements of each matrix. For the equation

, summing the diagonal entries yields the relationship $\operatorname{Tr}(A)+2\operatorname{Tr}(B) = 1-3+1 = -1$. Similarly, the trace of the equation

is found by summing its diagonal entries, resulting in $2-1+2 = 3$, which gives the relation $2\operatorname{Tr}(A)-\operatorname{Tr}(B) = 3$.

The resulting system of linear equations allows for the isolation of the individual trace values. Multiplying the equation $2\operatorname{Tr}(A)-\operatorname{Tr}(B) = 3$ by two results in $4\operatorname{Tr}(A)-2\operatorname{Tr}(B) = 6$. Adding this to the first equation, $\operatorname{Tr}(A)+2\operatorname{Tr}(B) = -1$, eliminates the trace of $B$, producing $5\operatorname{Tr}(A) = 5$, which simplifies to $\operatorname{Tr}(A) = 1$. Substituting this value into the expression $2\operatorname{Tr}(A)-\operatorname{Tr}(B) = 3$ gives $2(1)-\operatorname{Tr}(B) = 3$, leading to $\operatorname{Tr}(B) = -1$. The final value is calculated as $\operatorname{Tr}(A)-\operatorname{Tr}(B) = 1 - (-1) = 2$.

Determinants involving trigonometric identities are best managed by applying row or column operations that leverage the identity $\sin^2 x + \cos^2 x = 1$. By adding the second row to the first row, the elements in the first row become $1 + \sin^2 x + \cos^2 x$, which results in $2$ for the first two entries and $1$ for the third entry. This transformation simplifies the first row of the matrix to $[2, 2, 1]$.

Subtracting the second column from the first column then introduces two zeros in the first column, which makes evaluating the determinant significantly more straightforward. Expanding this determinant along the first column, the equation becomes

. Calculating the resulting $2 \times 2$ determinant yields $2(1 + 4\sin 2x) - 4\sin 2x = 0$, which simplifies to $2 + 8\sin 2x - 4\sin 2x = 0$, or $4\sin 2x = -2$.

This leads to the simplified equation $\sin 2x = -1/2$. Given the domain $0 < x < \pi$, the value of $2x$ must fall within the interval $0 < 2x < 2\pi$. Since the sine value is negative, the angle $2x$ must terminate in the third or fourth quadrant, specifically at $2x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ and $2x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. Dividing these values by $2$ provides the final solutions $x = \frac{7\pi}{12}$ and $x = \frac{11\pi}{12}$.

The key to solving this problem lies in recognizing that the constraint given for the rows of the matrix represents a specific property of matrix multiplication. If we multiply the matrix $A$ by a column vector containing all ones, denoted as

, each row of the resulting vector is the sum of the corresponding row elements of $A$. Since each row sum is 1, we obtain the relation $AX = X$.

Because this relationship holds for the matrix $A$, it extends naturally to any power of the matrix. By substituting $X$ with $AX$ in the expression for $A^2$, we find that $A^2X = A(AX) = AX = X$. Applying this iterative process one step further confirms that $A^3X = X$. This result indicates that the sum of the entries in each row of the matrix $A^3$ must equal 1. Given that there are three such rows in a $3 \times 3$ matrix, the total sum of all entries in $A^3$ is found by adding these three row sums, which results in a final value of 3.

The determinant of a diagonal matrix is calculated as the product of its diagonal entries, which simplifies the evaluation of the condition $\det(A^{2}-\frac{1}{2}I)=0$. Squaring the matrix $A$ results in:

$A^{2} = \begin{pmatrix} \sin^{2}\alpha & 0 \\ 0 & \sin^{2}\alpha \end{pmatrix}$

Subtracting $\frac{1}{2}I$ gives a matrix with diagonal entries $\sin^{2}\alpha-\frac{1}{2}$, and setting its determinant to zero leads to $(\sin^{2}\alpha-\frac{1}{2})^{2}=0$. Solving for $\alpha$ from $\sin^{2}\alpha = \frac{1}{2}$ yields $\sin\alpha = \pm\frac{1}{\sqrt{2}}$, which identifies $\frac{\pi}{4}$ as a valid solution.

The core principle here involves the relationship between a matrix and its determinant: the determinant of the square of a matrix is equal to the square of its determinant. Since the problem states that $A^{2}=I_{3}$, and the determinant of the identity matrix $I_{3}$ is $1$, we must have $|A|^{2}=1$. This establishes that the determinant of $A$ is either $1$ or $-1$.

Expanding the determinant of the matrix

yields the expression $|A| = x^{3}+y^{3}+z^{3}-3xyz$. Given the constraint $x+y+z > 0$, we select the positive value for the determinant, meaning the expression is equal to $1$. Setting this expansion equal to $1$, we get $x^{3}+y^{3}+z^{3}-3xyz = 1$. Substituting the given value $xyz = 2$ into this equation results in $x^{3}+y^{3}+z^{3}-3(2) = 1$, which simplifies to $x^{3}+y^{3}+z^{3} = 7$.

The properties of determinants allow for row and column operations that maintain the matrix's value while drastically simplifying the entries, providing the most efficient path to the solution. Applying the operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ isolates zeros in the third column, effectively reducing the order of the determinant. This transformation yields the following matrix:

$\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 4a+10 & 2 & 0 \end{vmatrix}$

With two zeros now in the third column, we expand along that column to find the result. The expansion simplifies to the calculation $1 \cdot (2(a+2) \cdot 2 - 1 \cdot (4a+10))$. Distributing the constants leads to $4(a+2) - (4a+10)$, which becomes $4a + 8 - 4a - 10$, ultimately resulting in a value of $-2$.

Simplifying a determinant through row transformations is an effective way to isolate terms and reduce complexity, particularly when trigonometric expressions are involved. Applying the operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$ converts the original matrix into a form that is much easier to evaluate:

Expanding this matrix yields $-2(\cos^{2} x) + 2(2 + 2 \cos 2x + \sin^{2} x)$, which simplifies further to $4 + 4 \cos 2x - 2(\cos^{2} x - \sin^{2} x)$. By applying the identity $\cos 2x = \cos^{2} x - \sin^{2} x$, the function reduces to $f(x) = 4 + 2\cos 2x$. Since the maximum value of the cosine function is 1, the maximum value of $f(x)$ is $4 + 2(1) = 6$.

Determinant properties allow for the simplification of rows and columns by applying linear operations without changing the value of the matrix. Given that $a, b, c,$ and $d$ are in arithmetic progression with a common difference $\lambda$, we know $b-a = c-b = d-c = \lambda$, as well as $a-c = -2\lambda$ and $-b+d = 2\lambda$. Applying the column operation $C_2 \rightarrow C_2 - C_3$ results in a constant $\lambda$ for every element in the second column, which simplifies the original expression to a matrix where $\lambda$ can be factored out. Subsequent row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ transform the determinant into

. Expanding this determinant along the second column involves the cofactor $-1$, leading to the evaluation $-\lambda \times ((2\lambda-1)(2\lambda) - (4\lambda)(\lambda)) = 2$, which resolves to $-\lambda \times (-2\lambda) = 2\lambda^2$. Equating this to the original value gives $2\lambda^2 = 2$, confirming that $\lambda^2 = 1$.

A homogeneous system of linear equations, written as $Ax = 0$, consistently includes the trivial solution $(0, 0, 0)$ and admits infinitely many solutions only when the determinant of the coefficient matrix is zero. Whenever this determinant vanishes, the equations are linearly dependent and do not restrict variables to a single point, allowing for a continuum of non-trivial solutions.

The determinant of the system is given by:
$\Delta = \begin{vmatrix} 1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1 \end{vmatrix}$
Expanding this yields the expression $1 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma + 2 \cos \alpha \cos \beta \cos \gamma$. With the condition $\alpha + \beta + \gamma = 2\pi$, this trigonometric result simplifies to exactly zero. Because the determinant equals zero, the system represents dependent equations and possesses infinitely many solutions.

Systems of linear equations possess no solution when the coefficient matrix is singular-meaning its determinant is zero-and the equations form an inconsistent set of constraints. Computing the determinant of the coefficient matrix yields $D = 3-a$, which forces the condition $a=3$ for the system to lack a unique solution. When this value is substituted into the system, the sum of the first two equations results in the linear expression $3x+5y+9z=13$. Because the third equation is defined as $3x+5y+9z=b$, a contradiction arises unless $b=13$, meaning the system is inconsistent whenever $b \neq 13$.

Powers of a square matrix are determined through iterative multiplication, which allows us to equate the results with the linear combination $5I - 8P$. First, evaluate this expression by substituting the given matrices:
$5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - 8 \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 16 & -8 \\ 40 & -24 \end{bmatrix} = \begin{bmatrix} -11 & 8 \\ -40 & 29 \end{bmatrix}$
Now, calculate $P^2$ as the product of $P$ with itself:
$\begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$
Proceed to determine $P^3$ by multiplying $P^2$ with the original matrix $P$:
$\begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix}$
Since $P^6$ is equivalent to $P^3$ multiplied by itself, compute this final square:
$\begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix} = \begin{bmatrix} -11 & 8 \\ -40 & 29 \end{bmatrix}$
Because this result matches our initial evaluation of $5I - 8P$, the value of $n$ is confirmed as 6.

The behavior of matrix powers for an upper triangular matrix like $A$ follows a predictable pattern where
$A^{n} = \begin{bmatrix} 1 & n & \frac{n^{2}+n}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$
When computing the sum $M = A + A^{2} + \dots + A^{20}$, the matrices are added together element-wise. The three diagonal elements are all 1 in each power of $A$, contributing $20 \times 3 = 60$ to the grand total. The positions (1,2) and (2,3) each contain $n$ in $A^n$, so the sum of these entries across all 20 matrices is $2 \times \frac{20 \times 21}{2} = 420$. The top-right element (1,3) contains $\frac{r^{2}+r}{2}$ for each $A^r$, and summing this expression for $r$ from 1 to 20 results in 1540. Combining these sums gives a total value of 2020.

Commuting matrices rely on the property $AB=BA$, which imposes strict constraints on the elements of an arbitrary $3 \times 3$ matrix $B$. Defining $B$ with nine variables, calculating the product $AB$ essentially swaps the first two rows of $B$, while computing $BA$ swaps the first two columns. Equating the corresponding entries of these products yields the necessary relationships $d=b$, $e=a$, $f=c$, and $g=h$. These constraints restrict the matrix $B$ to the form

, which depends on five independent variables. Since each of these five variables can be independently chosen from the set $\{1, 2, 3, 4, 5\}$, there are $5^5$ possible matrices, resulting in 3125.

Matrix identities and the properties of invertible operators provide a direct path to evaluating the determinant. Starting from the two given equations $A^{5} = B^{5}$ and $A^{3} B^{2} = A^{2} B^{3}$, we can subtract them to create a common factor. This subtraction results in $A^{5} - A^{3} B^{2} = B^{5} - A^{2} B^{3}$, which can be rewritten through careful grouping to isolate the structure of the matrices involved.

Factoring $A^{3}$ from the left side and $-B^{3}$ from the right side reveals:

$A^{3}(A^{2} - B^{2}) = -B^{3}(A^{2} - B^{2})$

Bringing all terms to one side yields the equation $(A^{3} + B^{3})(A^{2} - B^{2}) = 0$. Because the problem specifies that $(A^{2} - B^{2})$ is an invertible matrix, multiplying by its inverse on the right effectively cancels it out, leaving the identity $A^{3} + B^{3} = 0$. Consequently, the determinant of the zero matrix is 0.