Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

When a matrix satisfies the condition $A^2 = I$, the determinant of $A^2$ must equal the determinant of the identity matrix, which is $1$. Because the determinant of a product is the product of the determinants, $(\det(A))^2 = 1$, implying that $\det(A)$ must be either $1$ or $-1$. A $2 \times 2$ matrix with $A^2 = I$ has eigenvalues of $1$ or $-1$; if $A$ is neither $I$ nor $-I$, it must possess both $1$ and $-1$ as eigenvalues, forcing the determinant to be $1 \times (-1) = -1$. This confirms the validity of Statement 1.

Since the trace of a matrix is the sum of its eigenvalues, a matrix that is not $I$ or $-I$ must have a trace equal to $1 + (-1) = 0$. The second statement claims that the trace of such a matrix cannot be zero, which is directly contradicted by this result. This proves that Statement 2 is false.

A system of linear homogeneous equations possesses non-trivial solutions if and only if the determinant of its coefficient matrix is zero. Arranging the given equations into the form $-x + cy + bz = 0$, $cx - y + az = 0$, and $bx + ay - z = 0$ allows us to express this condition as:

Evaluating this determinant results in $-1(1 - a^2) - c(-c - ab) + b(ac + b) = 0$, which expands to $-1 + a^2 + c^2 + abc + abc + b^2 = 0$. This simplifies to the identity $a^2 + b^2 + c^2 + 2abc - 1 = 0$, confirming that $a^2 + b^2 + c^2 + 2abc$ must equal $1$.

The determinant of an upper triangular matrix is the product of its diagonal entries, which leads to $|A| = 25\alpha$. Given the property that the determinant of a squared matrix is the square of its determinant, the condition $|A^2|=25$ implies $|A|^2=25$. Substituting $25\alpha$ for $|A|$ into this equation gives $(25\alpha)^2 = 25$, and simplifying this results in $|\alpha| = \frac{1}{5}$.

When calculating a determinant, row operations allow us to simplify the matrix into an upper triangular form where all elements below the main diagonal are zero. By subtracting the first row from both the second and third rows, the matrix structure becomes significantly simpler. Specifically, subtracting the first row from the second row yields $(0, x, 0)$, and subtracting the first row from the third row yields $(0, 0, y)$, resulting in the determinant

. Since the determinant of an upper triangular matrix is the product of its diagonal entries, the value is $D = 1 \cdot x \cdot y = xy$, which confirms that the expression is divisible by both $x$ and $y$.

Matrix multiplication is generally non-commutative, so expanding the product $(A - B)(A + B)$ yields $A^2 + AB - BA - B^2$ rather than the simple $A^2 - B^2$ observed in scalar algebra. Setting this expanded form equal to $A^2 - B^2$ as stipulated in the problem results in the equation $A^2 + AB - BA - B^2 = A^2 - B^2$. Subtracting $A^2$ and adding $B^2$ to both sides cancels these terms out, leaving $AB - BA = 0$, which confirms that $AB = BA$.

Two matrices commute if their product remains unchanged regardless of the order of multiplication, requiring the equality $AB=BA$ to hold. Performing the matrix multiplications gives

and

. Comparing corresponding entries shows that $2b = 2a$ and $3a = 3b$, both of which simplify to the single condition $a=b$. Because the variables $a$ and $b$ must be natural numbers, any choice where $a$ equals $b$ defines a valid matrix $B$, and since there are infinitely many natural numbers, there exist infinitely many such matrices.

Simplify the determinant by performing the column operation $C_1 \rightarrow C_1 + C_2 + C_3$. This transforms every entry in the first column into $1 + x(a^2 + b^2 + c^2 + 2)$, and since the condition $a^2 + b^2 + c^2 = -2$ makes this expression equal to $1$, the entire first column becomes a column of ones.

Apply the row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$ to isolate the variable $x$. This reduces the matrix to

. Expanding this determinant yields $f(x) = (x-1)^2$, which is a quadratic expression. Consequently, the degree of the polynomial is 2.

The relationship between a matrix and its inverse is defined by the property $A A^{-1} = I$, which allows us to isolate $A^{-1}$ from any polynomial equation involving $A$. Starting with the equation $A^2 - A + I = 0$, we multiply every term by $A^{-1}$ to obtain $A - I + A^{-1} = 0$. Rearranging these terms to solve for the inverse directly yields $A^{-1} = I - A$.

A system of linear equations has the potential to possess infinite solutions only when the determinant of the coefficient matrix is equal to zero. For the provided equations, the determinant is defined as:

$\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha)$

Expanding and simplifying the determinant leads to $(\alpha - 1)(\alpha^2 + \alpha - 2)$, which factors further into $(\alpha - 1)^2(\alpha + 2)$. Setting the determinant to zero reveals potential values at $\alpha = 1$ and $\alpha = -2$. Because the value $\alpha = 1$ is excluded for this specific system, the condition for infinite solutions is met exclusively at $\alpha = -2$.

Since the terms $a_n, a_{n+1}, a_{n+2}, \ldots$ are in a geometric progression, we can express each term as $a_n = a r^{n-1}$, which implies their logarithms satisfy $\log(a_n) = \log a + (n-1)\log r$. Because of this linear dependence on the index, the differences between consecutive logarithmic terms are constant, specifically $\log(a_{k+1}) - \log(a_k) = \log r$. By performing the column operations $C_2 - C_1$ and $C_3 - C_2$, every entry in the second and third columns reduces to the constant $\log r$. Consequently, the determinant features two identical columns, which dictates that its total value is $0$.

Calculating the square of the matrix

requires performing matrix multiplication of $A$ with itself. By computing the dot product of the rows of the first matrix with the columns of the second, we determine the resulting matrix elements:

$\begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0(0)+0(0)+(-1)(-1) & 0(0)+0(-1)+(-1)(0) & 0(-1)+0(0)+(-1)(0) \\ 0(0)+(-1)(0)+0(-1) & 0(0)+(-1)(-1)+0(0) & 0(-1)+(-1)(0)+0(0) \\ -1(0)+0(0)+0(-1) & -1(0)+0(-1)+0(0) & -1(-1)+0(0)+0(0) \end{pmatrix}$

$= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I$

Because this product yields the identity matrix, the relationship $A^2 = I$ is established. Since the matrix is neither the zero matrix nor equal to $(-1)I$, and the determinant is non-zero ensuring the inverse exists, the equality $A^2 = I$ represents the only valid statement among the given options.

By definition, the product of a matrix and its inverse is the identity matrix, meaning $AB = I$. Multiplying both sides of this equation by 10 simplifies the calculation significantly, resulting in $A(10B) = 10I$. Since $10I$ is a diagonal matrix with 10s along the main diagonal and zeros elsewhere, the product of the first row of $A$ and the third column of $10B$ must equal the element in the first row and third column of $10I$, which is 0. Computing this dot product yields $(1)(2) + (-1)(\alpha) + (1)(3) = 2 - \alpha + 3$, which simplifies to $5 - \alpha$. Setting this result equal to 0 provides $5 - \alpha = 0$, confirming that $\alpha = 5$.

The terms of a geometric progression follow the form $a_k = a_1 r^{k-1}$, which means their logarithms are defined by $\log a_k = \log a_1 + (k-1) \log r$. By substituting these logarithmic expressions into the matrix, every entry takes the form of a linear arithmetic sequence. Specifically, the matrix represents columns where each term is generated by adding a constant increment, $\log r$, to the previous terms in a way that establishes a linear relationship between the columns.

Because the property of a geometric progression dictates that $2 \log a_{k+1} = \log a_k + \log a_{k+2}$, every element in the second column is exactly the arithmetic mean of the corresponding elements in the first and third columns. This signifies that the second column can be expressed as $C_2 = \frac{1}{2}(C_1 + C_3)$. In linear algebra, whenever one column of a determinant is a linear combination of the other columns, the rows and columns are linearly dependent, which forces the value of the entire determinant to be $0$.

Multiplying a matrix by itself requires calculating the dot product of rows from the first matrix and columns from the second, where each entry in the product matrix is the sum of these row-column products. By squaring

, the element in the first row and first column results in $a^2 + b^2$, while the element in the first row and second column is $(a \cdot b) + (b \cdot a) = 2ab$. This reveals the resulting matrix as

, meaning that the scalar values match $\alpha = a^{2+}b^2$ and $\beta = 2ab$.

The cyclic nature of complex cube roots of unity, defined by the identity $\omega^3 = 1$, allows for the simplification of this determinant. Expanding the matrix along the first row produces the expression $1(\omega^{3n} - 1) - \omega^n(\omega^{2n} - \omega^{2n}) + \omega^{2n}(\omega^n - \omega^{4n})$. Because $\omega^{3n}$ equals $1$ and $\omega^{6n}$ also simplifies to $1$, the expanded terms become $\omega^{3n} - 1 - 0 + \omega^{3n} - \omega^{6n}$. When these values are substituted, the arithmetic reduces to $1 - 1 + 1 - 1$, which results in zero.

A homogeneous system of linear equations possesses a non-zero solution if and only if the determinant of its coefficient matrix equals zero. Applying this principle to the system, the determinant condition is

.

Performing the column operation $C_2 \to C_2 - 2C_3$ simplifies the middle column, yielding

. Applying the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$ transforms the matrix into

.

Expanding this determinant along the first column produces $b(c-b) - (b-a)(2c-b) = 0$. Simplifying this expression results in $bc - b^2 - (2bc - b^2 - 2ac + ab) = 0$, which further reduces to $2ac = ab + bc$. Dividing the entire equation by $abc$ gives $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$, which confirms that $a, b,$ and $c$ are in harmonic progression.

Determinants exhibit the property of linearity, allowing the third column of the matrix to be separated into two distinct matrices. This split yields the equation

. By rearranging columns in the first determinant and factoring $a$, $b$, and $c$ from the rows of the second, the equation transforms into

. Because the vectors $(1, a, a^2)$, $(1, b, b^2)$, and $(1, c, c^2)$ are non-coplanar, their scalar triple product must be non-zero, permitting division by the determinant term. Consequently, the remaining factor $1+abc$ must be zero, identifying $abc$ as $-1$.

Simplifying a determinant relies on performing row or column operations to introduce zeros, which allows for a straightforward expansion. By applying the row transformation $R_3 \rightarrow R_3 - (x R_1 + R_2)$, the first two entries of the third row become zero because $ax+b - (x \cdot a + b) = 0$ and $bx+c - (x \cdot b + c) = 0$. The resulting third row contains only the entry $-(ax^2 + 2bx + c)$ in the third column.

Expanding this matrix along the third row yields the product $-(ax^2 + 2bx + c) \cdot (ac - b^2)$. Simplifying this expression leads to the final form:
$(b^2 - ac)(ax^2 + 2bx + c)$
Given that the discriminant of the quadratic is negative, the term $(b^2 - ac)$ must be negative. Since $a > 0$ and the discriminant is negative, the quadratic expression $ax^2 + 2bx + c$ is positive for all real $x$. Therefore, the product of these factors results in a negative value.

The determinant of the double adjoint of a square matrix $A$ of order $n$ is given by the identity $|\operatorname{adj}(\operatorname{adj}(A))| = |A|^{(n-1)^2}$. For this specific three-by-three matrix, the relationship simplifies to $|\operatorname{adj}(\operatorname{adj}(A))| = |A|^4$.

Calculating the determinant of the provided matrix reveals its value by factoring out 14 from each of its three rows, resulting in $14^3$ multiplied by the determinant of

. Since the determinant of this matrix evaluates to 14, the overall value is $14^4$. Equating $|A|^4$ to $14^4$ demonstrates that the positive value of $|A|$ is 14.

The matrix $A$ is defined by its action on three linearly independent vectors, which identify its eigenvalues as 2, 4, and 2. Because $A$ operates on the vectors

,

, and

by scaling them, these values are the characteristic properties of the transformation. When considering the system involving $(A-3I)$, we are essentially shifting the spectrum of $A$ by subtracting 3 from each eigenvalue. This transformation results in eigenvalues for the new matrix of $2-3=-1$, $4-3=1$, and $2-3=-1$.

A linear system $MX=B$ possesses a unique solution if and only if the matrix $M$ is invertible, which occurs when its determinant is non-zero. The determinant of a square matrix is equivalent to the product of its eigenvalues. For the matrix $(A-3I)$, this product is $(-1) \times 1 \times (-1) = 1$. Since the determinant is non-zero, the matrix $(A-3I)$ is non-singular and invertible. Consequently, the equation

is guaranteed to yield a unique solution for the variables $x$, $y$, and $z$.