Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of linear equations is inconsistent when its coefficient determinant, $\Delta$, is zero, provided that at least one of the associated determinants-$\Delta_x$, $\Delta_y$, or $\Delta_z$-is non-zero. The coefficient matrix determinant is

, which evaluates to $3(4+4) + 2(-2+2) - k(4+4) = 24 - 8k$. Setting $\Delta = 0$ yields $24 - 8k = 0$, requiring $k = 3$ for potential inconsistency.

With $k=3$ established, we evaluate the remaining determinants to ensure they do not all vanish. The determinant $\Delta_x$ becomes

, which simplifies to $32 - 40m$, while $\Delta_z$ similarly simplifies to $32 - 40m$ and $\Delta_y$ consistently evaluates to $0$. For the system to be inconsistent, at least one of these determinants must be non-zero, meaning $32 - 40m \neq 0$, which leads to the condition $m \neq \frac{4}{5}$.

A central principle in matrix algebra states that the determinant of any skew-symmetric matrix of odd order is always zero. To determine the nature of the system defined by $P = A^{2} B^{2}-B^{2} A^{2}$, we calculate the transpose $P^{T} = (A^{2} B^{2}-B^{2} A^{2})^{T}$. Using the properties of transposes, this becomes $P^{T} = (B^{2})^{T}(A^{2})^{T}-(A^{2})^{T}(B^{2})^{T}$. Since $A$ is symmetric, $A^{T} = A$, which implies that $A^{2}$ is symmetric, and since $B$ is skew-symmetric, $B^{T} = -B$, which leads to $(B^{2})^{T} = (-B)(-B) = B^{2}$, meaning $B^{2}$ is also symmetric. Thus, the expression simplifies to $P^{T} = B^{2} A^{2}-A^{2} B^{2} = -P$, confirming that $P$ is skew-symmetric. As $P$ is a $3 \times 3$ skew-symmetric matrix, its determinant is necessarily zero, ensuring that the system $PX = 0$ must possess infinitely many solutions.

For a system of three linear equations to possess infinitely many solutions, the determinant of the coefficient matrix must be equal to zero, as this condition signifies that the equations are linearly dependent and do not represent independent planes intersecting at a single unique point. To determine the necessary value of $k$, set the determinant of the system's coefficients to zero:

Expanding this determinant along the first row provides the equation $3(2k - (-15)) - (-1)(1k - (-18)) + 4(5 - 12) = 0$. Simplifying the terms within the parentheses results in $3(2k + 15) + (k + 18) + 4(-7) = 0$, which expands further to $6k + 45 + k + 18 - 28 = 0$. Combining the like terms yields $7k + 35 = 0$, which isolates to $k = -5$.

When variables form an arithmetic progression, we can express each term using the first term and the common difference, such that $y = x + d$ and $z = x + 2d$. Substituting these expressions into the determinant allows us to simplify the matrix using row operations, specifically subtracting the first row from the second and the second row from the third.

Performing the transformation $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{2}$ yields the matrix

Subtracting the second row from the third row results in

Expanding this determinant along the third row gives the product $-(k - 6\sqrt{2})(3d - x) = 0$. Since the problem explicitly states that $x \neq 3d$, the term $(3d - x)$ cannot be zero, which leaves $k - 6\sqrt{2} = 0$ as the only possible condition. Solving this leads to $k = 6\sqrt{2}$, and squaring both sides confirms that $k^2 = 72$.

Understanding how determinants respond to matrix transformations simplifies this problem significantly. For any $3 \times 3$ matrix $A$, scaling the entire matrix by 2 increases the determinant by a factor of $2^3$, meaning $\det(2A) = 8 \det(A)$. Given that $\det(A) = 4$, calculating this results in $\det(2A) = 8 \times 4 = 32$. The row operation $R_2 \rightarrow 2R_2 + 5R_3$ is then applied to this matrix.

This operation effectively consists of two parts: scaling the second row by 2 and adding a scalar multiple of the third row to it. Adding a multiple of one row to another leaves the determinant unchanged, while scaling a single row by 2 multiplies the entire determinant by 2. Applying this logic to the previous determinant of 32, the final value becomes $2 \times 32 = 64$. Alternatively, constructing the matrix $B$ shows its rows are $[2a, 2b, 2c]$, $[4d+10g, 4e+10h, 4f+10i]$, and $[2g, 2h, 2i]$. Removing the dependency on the third row confirms that $\det(B)$ simplifies to $2 \times 4 \times 2 \times \det(A)$, which is $16 \times 4 = 64$.

A system of linear equations possesses infinitely many solutions when the determinant of its coefficient matrix is zero, and the determinants formed by replacing any column of coefficients with the constant terms are also zero. This condition indicates that the equations are dependent, thereby preventing the existence of a single, unique solution.

Evaluating the determinant $\Delta_y$ requires replacing the $y$ coefficients with the constant terms $[1, 2, 3]^T$ from the system of equations $kx + y + 2z = 1$, $3x - y - 2z = 2$, and $-2x - 2y - 4z = 3$. This creates the following determinant:
$\Delta_y = \begin{vmatrix} k & 1 & 2 \\ 3 & 2 & -2 \\ -2 & 3 & -4 \end{vmatrix} = 0$
Expanding this across the first row yields the equation $k(-8+6) - 1(-12-4) + 2(9+4) = 0$. Simplifying this results in $-2k + 16 + 26 = 0$, which further simplifies to $-2k + 42 = 0$. Solving for $k$ gives $2k = 42$, identifying $k = 21$ as the required value.

The property $A A^T = I$ defines an orthogonal matrix, requiring the product of the matrix and its transpose to yield the identity matrix. Computing the product of

and its transpose

results in the matrix

. By equating this product to the identity matrix

, we find from the top-left element that $1+\alpha^2 = 1$, which dictates that $\alpha = 0$. Substituting $\alpha = 0$ into the bottom-right element, we get $0^2 + \beta^2 = 1$, implying $\beta^2 = 1$. Consequently, the expression $\alpha^{4+}\beta^4$ simplifies to $0^4 + 1^2 = 1$.

For any invertible square matrix $A$, the inverse is calculated as $A^{-1} = \frac{1}{|A|} \operatorname{adj} A$. Applied to

, the determinant is $|A| = (2 \times 4) - (2 \times 9) = 8 - 18 = -10$, and the adjugate is

.

Multiplying this inverse by 10 yields

, which simplifies to

. Comparing this result to the expression $A-6I$ reveals that

. Since both calculations result in the same matrix, the equivalence is confirmed.

The determinant of a matrix scales multiplicatively when its entries are modified by row and column factors. Given the relationship $b_{ij} = 3^{i+j-2} a_{ji}$, the matrix $B$ can be expressed as the product $D A^T D$ where $D = \text{diag}(1, 3, 9)$, which leads to the determinant relation $|B| = |D|^2 |A|$. Substituting the determinant of the diagonal matrix, $|D| = 1 \times 3 \times 9 = 27$, we obtain the equation:
$|B| = 27^2 |A| = 729 |A|$
With $|B| = 81$, this relationship becomes $81 = 729 |A|$, which simplifies to $|A| = \frac{1}{9}$.

The matrix $P$ possesses a determinant of zero, which signifies that the homogeneous system $PX=0$ has infinitely many solutions forming a line in three-dimensional space that passes through the origin. Expressing these solutions in parametric form relative to a scalar $\lambda$ yields $x = - \frac{11\lambda}{2}$, $y = \lambda$, and $z = \frac{7\lambda}{2}$, which satisfies every equation in the system. When this line is constrained by the unit sphere equation $x^2 + y^2 + z^2 = 1$, substituting the parametric expressions results in the quadratic form $\left( - \frac{11\lambda}{2} \right)^2 + \lambda^2 + \left( \frac{7\lambda}{2} \right)^2 = 1$. Solving this quadratic equation for $\lambda$ provides two distinct non-zero values, indicating that the line intersects the unit sphere at exactly two points.

Infinitely many solutions exist when the determinant of the coefficient matrix equals zero, along with the determinants of the matrices obtained by substituting the constant column into each variable's column. Finding the determinant of the coefficient matrix

gives $1(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 2\lambda - 9 = 0$, which results in $\lambda = 9/2$.

Using this value for $\lambda$, the condition for the $x$-variable determinant becomes

. Expanding this determinant yields $2(18 + 2) - 1(27 + \mu) + 1(12 - 4\mu) = 25 - 5\mu = 0$, confirming that $\mu = 5$. Substituting these values into $2\lambda + \mu$ results in $2(9/2) + 5 = 14$, which matches the condition provided.

An orthogonal matrix, such as the one defined here, requires that the sum of the squares of its row elements equals 1 and the sum of the products of elements from any two different rows equals 0. Applying these properties to the given matrix

results in the relations $a^2 + b^2 + c^2 = 1$ and $ab + bc + ca = 0$. By evaluating the expansion $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$, we find that $(a+b+c)^2 = 1$, which establishes that $a+b+c$ must be $1$ or $-1$. Substituting these figures into the factorization $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab + bc + ca))$ allows us to set $2 - 3abc = (\pm 1)(1 - 0)$, which simplifies to $3abc = 2 \mp 1$ and provides potential solutions for $abc$ as $1/3$ or $1$.

Applying elementary row operations is the most efficient strategy to simplify a determinant containing polynomial entries. By replacing the second row with the difference between the second and first rows, and the third row with the difference between the third and second rows, the matrix simplifies significantly. After performing these operations, the second row becomes $(x-1, x-1, x-1)$ and the third row becomes $(x-2, 2(x-2), 6(x-2))$. Because these rows now contain shared factors, we can factor out $(x-1)$ and $(x-2)$ to express the determinant as:

$\Delta = (x-1)(x-2) \begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ 1 & 1 & 1 \\ 1 & 2 & 6 \end{vmatrix}$

Evaluating this reduced determinant results in the product $-3(x-1)^{2}(x-2)$. Expanding this product creates the polynomial $-3x^{3}+12x^{2}-15x+6$. By matching this result to the standard form $Ax^{3}+Bx^{2}+Cx+D$, we identify the coefficients $B=12$ and $C=-15$. Summing these values yields $12-15$, which results in $-3$.

A homogeneous system of linear equations has a non-trivial solution if and only if the determinant of its coefficient matrix equals zero, which indicates that the equations are linearly dependent. For the matrix

, setting the determinant to zero involves calculating $1(9 - k^2) - 1(3 - 3k^2) + 3(1 - 9) = 0$, which simplifies to $2k^2 - 18 = 0$ and shows that $k^2 = 9$.

Substituting $k^2 = 9$ into the first and third equations gives $x + y + 3z = 0$ and $3x + y + 3z = 0$. Subtracting the first equation from the third reveals $2x = 0$, meaning $x$ must be $0$. Using $x = 0$ in the equation $x + y + 3z = 0$ results in $y + 3z = 0$, which implies that $\frac{y}{z} = -3$. Consequently, $x + (\frac{y}{z})$ evaluates to $0 + (-3)$, which is $-3$.

The system of equations $Ax_1 = b_1$, $Ax_2 = b_2$, and $Ax_3 = b_3$ is equivalent to the single matrix equation $AX = B$, where $X$ is the matrix formed by the column vectors $x_1, x_2, x_3$ and $B$ is the matrix formed by $b_1, b_2, b_3$. By the property that the determinant of a product is the product of the individual determinants, $|A||X| = |B|$, which means $|A| = |B| / |X|$. Constructing the matrix $X$ yields:
$X = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}$
This matrix is lower triangular, so its determinant is the product of its diagonal elements, which is $1 \times 2 \times 1 = 2$. Similarly, constructing the matrix $B$ gives:
$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
This is a diagonal matrix, and its determinant is the product of its diagonal elements, $1 \times 2 \times 2 = 4$. Dividing the determinant of $B$ by the determinant of $X$ results in $4 / 2 = 2$.

A system of linear equations possesses infinitely many solutions when the determinant of its coefficient matrix is zero and the system remains consistent, requiring the determinants of the associated augmented matrices to vanish as well. Calculating the determinant of the coefficient matrix gives:
$\begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{vmatrix} = 1(a+7) + 2(2a-1) + 3(-14-1) = 5a - 40$
Setting this determinant to zero gives $5a - 40 = 0$, which results in $a = 8$. Consistency requires that the determinant formed by replacing the first column with the constants, known as $D_1$, must also equal zero:
$\begin{vmatrix} 9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & 8 \end{vmatrix} = 9(15) + 2(8b-24) + 3(-7b-24) = 135 + 16b - 48 - 21b - 72 = -5b + 15 = 0$
The equation $-5b + 15 = 0$ provides $b = 3$. Consequently, finding the difference $a - b$ yields $8 - 3 = 5$.

For a system of linear equations to have no solution, the main determinant of the coefficient matrix must be zero, while at least one of the determinants associated with the variables, such as $D_x$, must be non-zero to indicate inconsistency. Calculating the coefficient determinant results in the quadratic equation $-2\lambda^2 + \lambda + 1 = 0$, which yields two distinct solutions for the parameter, specifically $\lambda = 1$ and $\lambda = -1/2$. Substituting these values into the expression for $D_x$ yields $2(1+\lambda)$, a value that is non-zero for both calculated roots. Because the system is inconsistent for both of these values, the set $S$ consists of exactly two elements.

The condition for a system of linear equations to possess infinitely many solutions is that the determinant of the coefficient matrix must be zero, and the determinants formed by replacing any column of the coefficient matrix with the column of constants must also be zero. Evaluating the determinant $D$ for the system $x+y+z=2$, $x+2y+3z=5$, and $x+3y+\lambda z=\mu$ provides the necessary equation to solve for $\lambda$:

$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 0$

Simplifying this expression yields $2\lambda - 9 - \lambda + 3 + 1 = 0$, which results in $\lambda - 5 = 0$, confirming that $\lambda = 5$. Applying this value to the determinant $D_1$, which replaces the first column with the constants, ensures the remaining condition for consistency:

$D_1 = \begin{vmatrix} 2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5 \end{vmatrix} = 2(10 - 9) - 1(25 - 3\mu) + 1(15 - 2\mu) = 0$

Expanding this determinant leads to $2 - 25 + 3\mu + 15 - 2\mu = 0$, which simplifies to $\mu - 8 = 0$, resulting in $\mu = 8$.

For a system of homogeneous linear equations to possess non-zero solutions, the determinant of its coefficient matrix must be equal to zero. Constructing the matrix from the given equations leads to

. Applying row operations by subtracting the first row from both the second and third rows simplifies the expression to $2((3b - 2a)(c - a) - (b - a)(4c - 2a)) = 0$. Expanding the terms within the equation yields $2ac - ab - bc = 0$, which can be rearranged to $2ac = ab + bc$. Dividing the entire equation by $abc$ results in $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$, establishing that the values $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ form an arithmetic progression.

Solving a system of linear equations relies on reducing the variables to express the coordinates in terms of a single parameter, which reveals the geometry of the solution space. By inspecting the given equations, one can eliminate $x$ and $y$ by noting that adding twice the first equation to the second equation yields $11z = 0$, which simplifies to $z = 0$. With $z$ eliminated, the first equation reduces to $x = 2y$, establishing that every valid triple must take the form $(2y, y, 0)$.

Substituting this relationship into the inequality $15 \le x^{2}+y^{2}+z^{2} \le 150$ results in the following constraint:
$15 \le (2y)^{2} + y^{2} + 0^{2} \le 150$
This simplifies to $15 \le 5y^{2} \le 150$, or equivalently $3 \le y^{2} \le 30$. Because $y$ must be an integer, its possible squares are $4, 9, 16,$ and $25$. These values correspond to $y \in \{2, 3, 4, 5\}$ and $y \in \{-2, -3, -4, -5\}$, which generate 8 distinct integer triples $(x, y, z)$ that satisfy the given conditions.