Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

A system of homogeneous linear equations possesses infinitely many solutions if and only if the determinant of its coefficient matrix equals zero. Evaluating the determinant for the given system yields the equation $2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) = 0$, which simplifies to the cubic polynomial equation $\lambda^3 + 4\lambda - 40 = 0$. The derivative of this polynomial, $3\lambda^2 + 4$, remains strictly positive for all real $\lambda$, which indicates that the function is strictly increasing across its entire domain. Since a strictly increasing cubic function crosses the x-axis at exactly one point, there is only one real value of $\lambda$ that satisfies the condition for infinitely many solutions.

Homogeneous systems of linear equations possess non-trivial solutions exclusively when the determinant of the coefficient matrix equals zero, as this condition ensures the equations are linearly dependent. The coefficient matrix for the given system is

, and setting its determinant to zero defines the requirement for non-trivial solutions. Expanding this determinant results in $1(\lambda + 1) - \lambda(-\lambda^2 + 1) - 1(\lambda + 1) = 0$, which simplifies to the cubic polynomial $\lambda^3 - \lambda = 0$. Factoring this expression as $\lambda(\lambda^2 - 1) = 0$ reveals three distinct roots for $\lambda$, specifically $0, 1,$ and $-1$. The existence of these three roots demonstrates that there are exactly three values of $\lambda$ for which the system yields non-trivial solutions.

Any square matrix satisfies the property that its product with its own adjoint equals its determinant multiplied by the identity matrix. Because the problem equates $A \operatorname{adj} A$ to $AA^T$, the matrix $|A|I$ must equal $AA^T$. The determinant of matrix $A$ is $(5a)(2) - (-b)(3)$, which simplifies to $10a+3b$. Multiplying this by the $2 \times 2$ identity matrix yields a diagonal matrix with $10a+3b$ on the main diagonal and zeros elsewhere.

Calculating $AA^T$ involves multiplying

by its transpose

, which results in
$\begin{bmatrix} 25a^{2+}b^2 & 15a-2b \\ 15a-2b & 13 \end{bmatrix}$
Comparing the off-diagonal entries to the zero entries in the diagonal matrix reveals that $15a-2b = 0$, meaning $b = 7.5a$. Equating the bottom-right entries gives $10a+3b = 13$. Substituting $b = 7.5a$ into this linear equation yields $10a+3(7.5a) = 32.5a = 13$, so $a = 0.4$. Consequently, $b$ is 3, and the value of $5a+b$ is 5.

The simplification of matrix powers involving conjugate expressions relies on the property of orthogonal matrices, where $P^TP = I$. Given the structure $Q = PAP^T$, raising the matrix $Q$ to a power $n$ results in $Q^n = (PAP^T)(PAP^T)...(PAP^T)$, which simplifies to $PA^nP^T$ because the identity $P^TP = I$ appears between each adjacent $A$ term. Consequently, the requested expression $P^TQ^{2015}P$ simplifies to $P^T(PA^{2015}P^T)P$, which further reduces to $(P^TP)A^{2015}(P^TP)$, and finally to $IA^{2015}I$, or simply $A^{2015}$.

For the matrix

, we can represent it as the sum of the identity matrix $I$ and a nilpotent matrix

. Because $N^2 = 0$, the binomial theorem states that $A^n = (I+N)^n = I + nN$. Applying this to the power $n=2015$, we get

, which yields

.

Statement I follows from the properties of matrix algebra by rearranging the equation $A^{2-}5A+7I=O$. Factoring out $A$ from $5A-A^2$ gives $A(5I-A)=7I$, and multiplying both sides by $A^{-1}$ confirms that $A^{-1}=\frac{1}{7}(5I-A)$.

The reduction of the polynomial $A^{3-}2A^{2-}3A+I$ is achieved by substituting the relation $A^2=5A-7I$ into the expression. This substitution converts the cubic term $A^3$ into $A(5A-7I)=5A^{2-}7A$. Replacing $A^2$ in the overall expression leads to $(5A^{2-}7A)-2A^{2-}3A+I$, which simplifies to $3A^{2-}10A+I$. Final substitution of $A^2=5A-7I$ yields $3(5A-7I)-10A+I$, which simplifies to $15A-21I-10A+I=5A-20I$, showing that the polynomial is equivalent to $5(A-4I)$.

The determinant of a product of matrices equals the product of their individual determinants, a property that allows for the simplification of complex matrix expressions. By factoring the matrix $A^{2014}$ out of the expression $A^{2016} - 2A^{2015} - A^{2014}$, we obtain the product $A^{2014}(A^2 - 2A - I)$, where $I$ represents the identity matrix. Since the determinant is multiplicative, the value we seek is the product of $\det(A^{2014})$ and $\det(A^2 - 2A - I)$.

Calculating the determinant of the given matrix

yields $(-4)(1) - (-1)(3)$, which equals $-4 + 3 = -1$. Consequently, $\det(A^{2014}) = (\det(A))^{2014} = (-1)^{2014} = 1$. Next, we focus on the term $A^2 - 2A - I$. Squaring matrix $A$ results in

. Subtracting $2A$, which is

, and $I$, which is

, from $A^2$ leaves us with the matrix

. Computing the determinant of this final matrix gives $(20)(-5) - (5)(-15) = -100 + 75 = -25$. Multiplying this result by the previously determined factor of $1$ confirms the value is $-25$.

A system of linear equations in the form $AX = \lambda X$ admits non-trivial solutions only when the determinant of the characteristic matrix $(A - \lambda I)$ equals zero. Organizing the given equations into a homogeneous system results in the requirement that the following determinant must vanish:

$\begin{vmatrix} 2-\lambda & -2 & 1 \\ 2 & -3-\lambda & 2 \\ -1 & 2 & -\lambda \end{vmatrix} = 0$

Evaluating this determinant produces the cubic equation $\lambda^3 + \lambda^2 - 5\lambda + 3 = 0$. Solving this polynomial reveals the values $\lambda = 1$ and $\lambda = 3$. Because the set of valid $\lambda$ values comprises exactly two distinct elements, the condition for a non-trivial solution is satisfied by a set containing two elements.

Starting with the definition $B = A^{-1}A'$, we evaluate the product $BB'$ by utilizing the property of transposes, $(XY)' = Y'X'$, which yields $BB' = (A^{-1}A')(A^{-1}A')' = A^{-1}A'(A')'(A^{-1})'$. By applying the identities $(A')' = A$ and $(A^{-1})' = (A')^{-1}$, the expression simplifies to $A^{-1}A'A(A')^{-1}$. Given that $A$ commutes with its transpose such that $A'A = AA'$, we can substitute this relationship into the equation to get $A^{-1}(AA')(A')^{-1}$. Finally, regrouping the terms through the associative property results in $(A^{-1}A)(A'(A')^{-1})$, which is the product of two identity matrices, confirming that $BB' = I$.

The relationship between a square matrix $A$ and its adjoint matrix $P$ is defined by the property $|\operatorname{adj}(A)| = |A|^{n-1}$, where $n$ represents the dimension of the matrix. Given that $A$ is a $3 \times 3$ matrix with a determinant of $4$, the determinant of its adjoint must be $4^{3-1}$, which calculates to $4^2$ or $16$.

By expanding the determinant of the provided matrix $P$ along the first row, we obtain $1(12 - 12) - \alpha(4 - 6) + 3(4 - 6)$. Simplifying these terms results in $0 - \alpha(-2) + 3(-2)$, which yields $2\alpha - 6$. Equating this result to the value derived from the adjoint property, we set $2\alpha - 6 = 16$. Adding 6 to both sides gives $2\alpha = 22$, which determines that $\alpha = 11$.

A system of linear equations of the form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ has no solution if the lines are parallel but distinct, which occurs when the ratios of the coefficients of the variables are equal to each other but not to the ratio of the constant terms, expressed as $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Applying this principle to the given system, we equate the variable ratios as $\frac{k+1}{k} = \frac{8}{k+3}$ and solve for $k$ by cross-multiplying to obtain $k^2 + 4k + 3 = 8k$, which simplifies to the quadratic equation $k^2 - 4k + 3 = 0$. Factoring this expression gives $(k-1)(k-3) = 0$, suggesting potential candidates of $k=1$ and $k=3$. Checking these values against the full condition, the substitution of $k=1$ results in coefficient ratios of $\frac{2}{1} = \frac{8}{4}$ and a constant ratio of $\frac{4}{2}$, meaning all three ratios are equal and the system is consistent with infinitely many solutions. Substituting $k=3$ yields coefficient ratios of $\frac{4}{3} = \frac{8}{6}$ and a constant ratio of $\frac{12}{8}$, which simplifies to $\frac{3}{2}$; since $\frac{4}{3} \neq \frac{3}{2}$, the system has no solution only for this specific value of $k$.

Leveraging the linearity property of matrix multiplication, we can combine the given expressions into the single equation

. Because $A$ is a lower triangular matrix, we can determine the components of $u_1 + u_2$, denoted $x, y,$ and $z$, efficiently via forward substitution using the system $x = 1$, $2x + y = 1$, and $3x + 2y + z = 0$. The first equation gives $x = 1$, which implies $2(1) + y = 1$ so $y = -1$. Substituting these values into the final equation, $3(1) + 2(-1) + z = 0$ results in $1 + z = 0$, leading to $z = -1$ and the final vector

.

The determinant of a matrix product is equal to the product of the individual determinants, so if the product of two matrices is the zero matrix, their determinant product must also vanish. Subtracting the given equation $P^2 Q = Q^2 P$ from $P^3 = Q^3$ yields $P^3 - P^2 Q = Q^3 - Q^2 P$, which can be rearranged and factored as $P^2(P - Q) + Q^2(P - Q) = 0$, resulting in $(P^2 + Q^2)(P - Q) = 0$. Given that $P \neq Q$, the condition $\det(P^2 + Q^2) \cdot \det(P - Q) = 0$ implies that $\det(P^2 + Q^2)$ must be $0$.

A system of homogeneous linear equations possesses non-zero solutions only when the determinant of its coefficient matrix is zero, which implies that the equations are linearly dependent. Setting the determinant of the coefficient matrix to zero results in the following equation:

Expanding this determinant gives $4(4-2) - k(k-2) + 2(2k-8) = 0$. Simplifying this expression leads to $8 - k^2 + 2k + 4k - 16 = 0$, which rearranges into the quadratic equation $k^2 - 6k + 8 = 0$. Factoring this quadratic into $(k-4)(k-2) = 0$ yields two distinct values for $k$, specifically $4$ and $2$, for which the system admits a non-trivial solution.

The symmetry of a matrix is defined by the property $A' = A$, and for symmetric matrices $A$ and $B$, the transpose $(A(BA))'$ expands to $(BA)'A' = A'B'A'$, which simplifies back to $A(BA)$ because $A' = A$ and $B' = B$. This proves the first statement is true independent of any commutative properties. Regarding the second statement, the transpose of $AB$ is $B'A'$, which equates to $BA$, meaning $AB$ is symmetric if and only if $AB = BA$. Since the first statement holds true without requiring commutativity, the second statement is a correct standalone fact but does not explain the symmetry established in the first statement.

The fundamental property $A^2=I$ establishes strict algebraic constraints on the entries of a matrix

. By computing the square of this matrix, the off-diagonal elements in the resulting product are $b(a+d)$ and $c(a+d)$. Given that both $b$ and $c$ are non-zero, the only way for these off-diagonal terms to equal zero, as required by the identity matrix, is for the term $(a+d)$ to be zero. Since the trace of a matrix is defined as the sum of its diagonal elements, $a+d=0$ confirms that the trace must indeed be zero.

The determinant of the matrix follows from the property $|A|^2 = |A^2| = |I| = 1$, which leaves the potential values for $|A|$ as either $1$ or $-1$. Using the specific entries, the determinant is calculated as $|A| = ad - bc$. Since the previous result established that $d = -a$, the determinant expression becomes $-a^2 - bc$. From the diagonal element of the identity equation $a^2 + bc = 1$, it is evident that $bc = 1 - a^2$. Substituting this into the determinant formula results in $|A| = -a^2 - (1 - a^2) = -1$. Because the determinant evaluates to $-1$ rather than $1$, the assertion that the determinant is $1$ is incorrect.

A matrix is classified as non-singular when its determinant $|A| \neq 0$, which depends on the specific arrangement of the four 1s and five 0s within the $3 \times 3$ grid. By fixing three 1s along the main diagonal, one can place the fourth 1 in any of the remaining 6 positions to create a distinct non-singular matrix. A parallel logic applies when fixing three 1s along the anti-diagonal and placing the fourth 1 in one of the other 6 empty spots, which identifies another group of 6 valid configurations. Since these two distinct sets alone provide more than enough combinations to exceed the specified range, the total number of such matrices is confirmed to be at least 7.

A system of linear equations fails to provide a unique solution when the determinant of its coefficient matrix, $D$, is exactly zero. Calculating the determinant for the provided system gives

, which immediately indicates that the system is either inconsistent or dependent. Evaluating the Cramer determinant

confirms a non-zero result. This discrepancy between a vanishing coefficient determinant and a non-vanishing Cramer determinant mathematically proves that there is no solution.

For a square matrix $A$ of order $n$, the double adjoint is governed by the identity $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$. Substituting $n = 2$ for the given matrix yields $\operatorname{adj}(\operatorname{adj} A) = |A|^{2-2} A = A$, confirming that the first statement is correct.

Regarding the determinant, a standard property states that the determinant of an adjoint matrix is $|\operatorname{adj} A| = |A|^{n-1}$. For a $2 \times 2$ matrix, this becomes $|\operatorname{adj} A| = |A|^{2-1} = |A|$, proving the second statement is true. Since the general relationship for the double adjoint is derived from the properties of the determinant of an adjoint matrix, the second statement acts as the theoretical foundation that validates the first.

Determinant properties allow the second matrix to be transformed into the same structure as the first, enabling the equation to be simplified. By taking the transpose and swapping columns, the expression becomes $[1 + (-1)^{n+2}]$ multiplied by the value of the first determinant, equal to zero. Computing this determinant results in $4b(a+c)$, and since the constraint $b(a+c) \neq 0$ ensures this value is non-zero, the multiplier $[1 + (-1)^{n+2}]$ must equal zero. This requires $(-1)^{n+2} = -1$, which is only possible if the exponent $n+2$ is odd, meaning $n$ must be an odd integer.

When a square matrix is composed entirely of integers, its adjugate matrix-formed from the matrix of cofactors-must also contain only integers because each cofactor is a determinant calculated from integer entries. Since the formula for the inverse is $A^{-1} = \frac{1}{\det A} \text{adj}(A)$, having a determinant of $\pm 1$ ensures that we are simply dividing an integer matrix by $\pm 1$, which guarantees that every entry in the resulting inverse matrix remains an integer.