Let P = [pij] and Q = [qij] be two square matrices of order 3 such that qij = 2(i + j

Q1 jee_main_2024_29_january_evening Properties of Determinants

Let

A = bmatrix 2 & 1 & 2 6 & 2 & 11 3 & 3 & 2 bmatrix

and

P = bmatrix 1 & 2 & 0 5 & 0 & 2 7 & 1 & 5 bmatrix

. The sum of the prime factors of

|P⁻¹AP - 2I|

is equal to

A. 26
B. 27
C. 66
D. 23

Solution

Related Formula

|P⁻¹AP - 2I| = |P⁻¹(A - 2I)P| = |P⁻¹| \cdot |A - 2I| \cdot |P| = |A - 2I|

Core Logic

Since $|P⁻¹| \cdot |P| = 1$ $|P^{-1}| \cdot |P| = 1$, the expression simplifies completely to the determinant of $A - 2I$ $A - 2I$.

First, let us construct the matrix $A - 2I$ $A - 2I$:

A - 2I = bmatrix 2-2 & 1 & 2 6 & 2-2 & 11 3 & 3 & 2-2 bmatrix = bmatrix 0 & 1 & 2 6 & 0 & 11 3 & 3 & 0 bmatrix

Now, evaluating the determinant:

|A - 2I| = 0(0 - 33) - 1(0 - 33) + 2(18 - 0) = 33 + 36 = 69

Step 1: Finding Prime Factors

The number obtained is $69$ $69$. Let us find its prime factorization:

69 = 3 \times 23

Both $3$ $3$ and $23$ $23$ are prime numbers. Their sum is:

Sum = 3 + 23 = 26

Pattern Recognition

Whenever you encounter a matrix expression of the form $P⁻¹AP - kI$ $P^{-1}AP - kI$, always factor out $P⁻¹$ $P^{-1}$ and $P$ $P$ to simplify it to $|A - kI|$ $|A - kI|$. This saves tremendous time over computing matrix multiplications.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q22 jee_main_2024_29_january_evening System of Linear Equations

Let for any three distinct consecutive terms

a, b, c

of an A.P, the lines

ax + by + c = 0

be concurrent at the point

P

and

Q (α, β)

be a point such that the system of equations

x + y + z = 6

,

2 x + 5 y + α z = β

and

x + 2y + 3z = 4

has infinitely many solutions. Then

(PQ)²

is equal to

Numerical Answer. Answer: 113 to 113

Solution

Related Formula

For infinite solutions of a system of equations, the main determinant $D$ $D$ and auxiliary determinants $D₁, D₂, D₃$ $D_1, D_2, D_3$ must all equal 0.

Core Logic

Since $a, b, c$ $a, b, c$ are in A.P., we have $2b = a + c a - 2b + c = 0$ $2b = a + c \implies a - 2b + c = 0$. Comparing this identity with the line equation $ax + by + c = 0$ $ax + by + c = 0$, we immediately find that the lines always pass through the fixed point configuration $(1, -2)$ $(1, -2)$. Thus, $P = (1, -2)$ $P = (1, -2)$.

Step 1: Evaluation of the Matrix for Infinite Solutions

Let us set the system determinant $D = 0$ $D = 0$:

D = bmatrix 1 & 1 & 1 2 & 5 & α 1 & 2 & 3 bmatrix = 0

1(15 - 2α) - 1(6 - α) + 1(4 - 5) = 0

15 - 2α - 6 + α - 1 = 0 8 - α = 0 α = 8

Now set $D₁ = 0$ $D_1 = 0$ by substituting columns:

D₁ = bmatrix 6 & 1 & 1 β & 5 & 8 4 & 2 & 3 bmatrix = 0

6(15 - 16) - 1(3β - 32) + 1(2β - 20) = 0

-6 - 3β + 32 + 2β - 20 = 0 6 - β = 0 β = 6

Thus, $Q = (8, 6)$ $Q = (8, 6)$.

Step 2: Distance Metric Calculation

Using the distance formula between $P(1, -2)$ $P(1, -2)$ and $Q(8, 6)$ $Q(8, 6)$:

(PQ)² = (8 - 1)² + (6 - (-2))² = 7² + 8² = 49 + 64 = 113

Pattern Recognition

A.P. coefficients inside a standard linear equation reveal a fixed point of concurrency by mapping matching coefficient components ( $1, -2, 1$ $1, -2, 1$).

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Straight Lines

Q19 jee_main_2024_27_jan_morning Matrix Multiplication

Consider the matrix

f(x) = bmatrix x & - x & 0 x & x & 0 0 & 0 & 1 bmatrix

Given below are two statements: Statement I:

f(-x)

is the inverse of the matrix

f(x)

Statement II:

f(x)f(y) = f(x+y)

. In the light of the above statements, choose the correct answer from the options given below

A. $Statement I is false but Statement II is true$
B. $Both Statement I and Statement II are false$
C. $Statement I is true but Statement II is false$
D. $Both Statement I and Statement II are true$

Solution

Related Formula

(-x) = x

(-x) = - x

(x+y) = x y + x y

(x+y) = x y - x y

Core Logic

Evaluate $f(-x)$ $f(-x)$:

f(-x) = bmatrix (-x) & - (-x) & 0 (-x) & (-x) & 0 0 & 0 & 1 bmatrix = bmatrix x & x & 0 - x & x & 0 0 & 0 & 1 bmatrix

Checking if $f(-x)$ $f(-x)$ is the inverse by evaluating $f(x) \cdot f(-x)$ $f(x) \cdot f(-x)$:

f(x) f(-x) = bmatrix ² x + ² x & x x - x x & 0 x x - x x & ² x + ² x & 0 0 & 0 & 1 bmatrix

f(x) f(-x) = bmatrix 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 bmatrix = I

Thus, Statement I is true.

Step 1: Checking Statement II

Evaluate the matrix multiplication $f(x) \cdot f(y)$ $f(x) \cdot f(y)$:

f(x) f(y) = bmatrix x & - x & 0 x & x & 0 0 & 0 & 1 bmatrix bmatrix y & - y & 0 y & y & 0 0 & 0 & 1 bmatrix

= bmatrix x y - x y & - x y - x y & 0 x y + x y & - x y + x y & 0 0 & 0 & 1 bmatrix

Apply standard trigonometric compound angle formulas:

= bmatrix (x+y) & - (x+y) & 0 (x+y) & (x+y) & 0 0 & 0 & 1 bmatrix = f(x+y)

Thus, Statement II is also true.

Step 2: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

This specific matrix represents a standard 2D rotation matrix embedded in 3D space. Rotation matrices naturally follow $R(x)R(y) = R(x+y)$ $R(x)R(y) = R(x+y)$ (additive property of angles) and their inverse is always obtained by negating the angle $R(-x) = R(x)⁻¹$ $R(-x) = R(x)^{-1}$.

Chapter Mix

Class 12 Maths: Matrices Class 11 Maths: Trigonometric Functions

Q29 jee_main_2024_27_jan_morning Matrix Inverse and Determinant

Let

A= bmatrix 2 & 0 & 1 1 & 1 & 0 1 & 0 & 1 bmatrix

,

B=[B₁, B₂, B₃]

, where

B₁, B₂, B₃

are column matrices, and

AB₁= bmatrix 1 0 0 bmatrix

,

AB₂= bmatrix 2 3 0 bmatrix

,

AB₃= bmatrix 3 2 1 bmatrix

. If

α=|B|

and

β

is the sum of all the diagonal elements of

B

, then

α³+β³

is equal to:

Numerical Answer. Answer: 28 to 28

Solution

Related Formula

$|AB| = |A||B|$ $|AB| = |A||B|$ Trace( $B$ $B$) = $Σ bᵢᵢ$ $\sum b_{ii}$

Core Logic

Since $B = [B₁, B₂, B₃]$ $B = [B_1, B_2, B_3]$, the matrix multiplication $AB$ $AB$ effectively applies $A$ $A$ to each column of $B$ $B$:

AB = [AB₁, AB₂, AB₃] = bmatrix 1 & 2 & 3 0 & 3 & 2 0 & 0 & 1 bmatrix

Let this target matrix be $C$ $C$. We know $AB = C$ $AB = C$. Thus, taking determinants on both sides: $|A| |B| = |C|$ $|A| |B| = |C|$

Step 1: Finding Determinants

Calculate determinant of $A$ $A$:

|A| = 2(1 - 0) - 0 + 1(0 - 1) = 2 - 1 = 1

Calculate determinant of $C$ $C$. Since $C$ $C$ is an upper triangular matrix, its determinant is simply the product of its main diagonal:

|C| = 1 \times 3 \times 1 = 3

Therefore, $1 \times |B| = 3 \Rightarrow |B| = 3$ $1 \times |B| = 3 \Rightarrow |B| = 3$. So $α = 3$ $\alpha = 3$.

Step 2: Finding Matrix B

To find $β$ $\beta$ (the trace of $B$ $B$), we must explicitly find $B = A⁻¹C$ $B = A^{-1}C$. Rather than finding the full inverse, manually solve $A Bᵢ = Cᵢ$ $A B_i = C_i$: For $B₁ = [x, y, z]^T$ $B_1 = [x, y, z]^T$: $2x+z=1$ $2x+z=1$, $x+y=0$ $x+y=0$, $x+z=0 \Rightarrow x=1, z=-1, y=-1$ $x+z=0 \Rightarrow x=1, z=-1, y=-1$. Thus $B₁ = [1, -1, -1]^T$ $B_1 = [1, -1, -1]^T$. For $B₂$ $B_2$: $2x+z=2$ $2x+z=2$, $x+y=3$ $x+y=3$, $x+z=0 \Rightarrow x=2, z=-2, y=1$ $x+z=0 \Rightarrow x=2, z=-2, y=1$. Thus $B₂ = [2, 1, -2]^T$ $B_2 = [2, 1, -2]^T$. For $B₃$ $B_3$: $2x+z=3$ $2x+z=3$, $x+y=2$ $x+y=2$, $x+z=1 \Rightarrow x=2, z=-1, y=0$ $x+z=1 \Rightarrow x=2, z=-1, y=0$. Thus $B₃ = [3, 0, -1]^T$ $B_3 = [3, 0, -1]^T$.

Step 3: Calculating Trace and Final Output

Constructing Matrix B:

B = bmatrix 1 & 2 & 3 -1 & 1 & 0 -1 & -2 & -1 bmatrix

The diagonal elements are $1, 1, -1$ $1, 1, -1$. Trace $β = 1 + 1 - 1 = 1$ $\beta = 1 + 1 - 1 = 1$.

Finally compute $α³ + β³$ $\alpha^3 + \beta^3$:

3³ + 1³ = 27 + 1 = 28

Pattern Recognition

When given $AXᵢ = Yᵢ$ $AX_i = Y_i$ for multiple columns, they collectively form $A X = Y$ $A X = Y$. Using $|A||X| = |Y|$ $|A||X| = |Y|$ bypasses full matrix inversion if you strictly need determinants. To grab the trace, solving equations systematically column-by-column is generally less error-prone than forming the full adjoint inverse matrix.

Chapter Mix

Class 12 Maths: Matrices

Q15 jee_main_2024_29_jan_morning Properties of Determinants

Let

A= bmatrix1&0&0 0&α&β 0&β&α bmatrix

and

|2A|³=2²¹

where

α, βin Z

, Then a value of

α

is

A. $3$ $3$
B. $5$ $5$
C. $17$ $17$
D. $9$ $9$

Solution

Related Formula

$|kA| = kⁿ |A|$ $|kA| = k^n |A|$ Where $A$ $A$ is an $n \times n$ $n \times n$ matrix, and $k$ $k$ is a scalar.

Core Logic

Find the determinant of the $3 \times 3$ $3 \times 3$ matrix $A$ $A$:

|A| = 1(α \cdot α - β \cdot β) - 0 + 0

|A| = α² - β²

Given the condition $|2A|³ = 2²¹$ $|2A|^3 = 2^{21}$. Since $A$ $A$ is a $3 \times 3$ $3 \times 3$ matrix, applying the scalar property $|kA| = k³|A|$ $|kA| = k^3|A|$:

|2A| = 2³|A| = 8|A|

Substitute this back into the original condition:

(2³|A|)³ = 2²¹

2⁹ |A|³ = 2²¹

|A|³ = 2²¹2⁹ = 2¹²

Taking the cube root of both sides: $|A| = 2⁴ = 16$ $|A| = 2^4 = 16$

Step 1: Solve the Diophantine Equation

We have:

α² - β² = 16

(α - β)(α + β) = 16

Since $α$ $\alpha$ and $β$ $\beta$ are integers, their sum and difference must also be integers. Also, $(α + β)$ $(\alpha + \beta)$ and $(α - β)$ $(\alpha - \beta)$ must share the same parity (both even or both odd) because their sum is $2α$ $2\alpha$ (an even number).

Since their product is 16, the only valid integer factor pairs of 16 that share the same parity are $(8, 2)$ $(8, 2)$ and $(-8, -2)$ $(-8, -2)$ and $(4, 4)$ $(4, 4)$ and $(-4, -4)$ $(-4, -4)$.

Case 1: $(α + β) = 8$ $(\alpha + \beta) = 8$ and $(α - β) = 2$ $(\alpha - \beta) = 2$ Adding them gives $2α = 10 \Rightarrow α = 5$ $2\alpha = 10 \Rightarrow \alpha = 5$. Thus $β = 3$ $\beta = 3$.

Case 2: $(α + β) = 4$ $(\alpha + \beta) = 4$ and $(α - β) = 4$ $(\alpha - \beta) = 4$ Adding them gives $2α = 8 \Rightarrow α = 4$ $2\alpha = 8 \Rightarrow \alpha = 4$. Thus $β = 0$ $\beta = 0$.

Looking at the options provided (3, 5, 17, 9), the value $α = 5$ $\alpha = 5$ is listed.

Pattern Recognition

Extracting scalar multipliers from determinants always depends on the dimension $n$ $n$ of the matrix. For Diophantine equations like $x² - y² = k$ $x^2 - y^2 = k$, factoring into $(x-y)(x+y)$ $(x-y)(x+y)$ and analyzing parity constraints restricts the solution space instantly.

Chapter Mix

Class 12 Mathematics: Determinants

Let P = [pᵢⱼ]$P = [p_{ij}]$ and Q = [qᵢⱼ]$Q = [q_{ij}]$ be two square matrices of order 3 such that qᵢⱼ = 2(i + j - 1) pᵢⱼ$q_{ij} = 2^{(i + j - 1)} p_{ij}$ and (Q) = 2¹⁰$\det(Q) = 2^{10}$. Then the value of (adj(adj P))$\det(\text{adj}(\text{adj } P))$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Factoring from Rows

Step 2: Factoring from Columns

Step 3: Finding |P| and Target Value

Pattern Recognition

Chapter Mix

Reference Study Guides

More Determinants Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Finding Prime Factors

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation of the Matrix for Infinite Solutions

Step 2: Distance Metric Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Checking Statement II

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Determinants

Step 2: Finding Matrix B

Step 3: Calculating Trace and Final Output

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve the Diophantine Equation

Pattern Recognition

Chapter Mix

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Let $P = [pᵢⱼ]$ $P = [p_{ij}]$ and $Q = [qᵢⱼ]$ $Q = [q_{ij}]$ be two square matrices of order 3 such that $qᵢⱼ = 2(i + j - 1) pᵢⱼ$ $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $(Q) = 2¹⁰$ $\det(Q) = 2^{10}$. Then the value of $(adj(adj P))$ $\det(\text{adj}(\text{adj } P))$ is: