For the matrices A=bmatrix3 & -4 1 & -1bmatrix and B=bmatrix-29 & 49 -13 & 18bmatrix, if (A15+B)bmatrixx ybmatrix=bmatrix0 0bmatrix, then among the following which one is true?

(provided N2 = 0, meaning N is nilpotent of order 2). Evaluate powers of matrix A to detect a pattern. Let N = bmatrix2 & -4 1 & -2bmatrix. Check N2: Since N2 = 0, An = I + nN. Given (A15 + B)bmatrixx ybmatrix = bmatrix0 0bmatrix: This yields a single independent equation: Checking the options, for x=11, y=2, 2(11) = 11(2) 22 = 22. Thus, Option 3 satisfies the condition. Whenever you need a high power of a 2x 2 matrix, check its trace and determinant. If |A-lambda I|=0 yields lambda=1,1, then A can be written as I+N where N is nilpotent. This bypasses lengthy matrix multiplication. Class 12 Maths: Matrices and Determinants

For the matrices A=bmatrix3 & -4 1 & -1bmatrix and B=bmatrix-29 & 49 -13 & 18bmatrix, if (A15+B)bmatrixx ybmatrix=bmatrix0 0bmatrix, then among the following which one is true?

(provided N2 = 0, meaning N is nilpotent of order 2). Evaluate powers of matrix A to detect a pattern. Let N = bmatrix2 & -4 1 & -2bmatrix. Check N2: Since N2 = 0, An = I + nN. Given (A15 + B)bmatrixx ybmatrix = bmatrix0 0bmatrix: This yields a single independent equation: Checking the options, for x=11, y=2, 2(11) = 11(2) 22 = 22. Thus, Option 3 satisfies the condition. Whenever you need a high power of a 2x 2 matrix, check its trace and determinant. If |A-lambda I|=0 yields lambda=1,1, then A

For the matrices A=bmatrix3 & -4 1 & -1bmatrix and B=bmatrix-29 & 49 -13 & 18bmatrix, if (A15+B)bmatrixx ybmatrix=bmatrix0 0bmatrix, then among the following which one is true?

(provided N2 = 0, meaning N is nilpotent of order 2). Evaluate powers of matrix A to detect a pattern. Let N = bmatrix2 & -4 1 & -2bmatrix. Check N2: Since N2 = 0, An = I + nN. Given (A15 + B)bmatrixx ybmatrix = bmatrix0 0bmatrix: This yields a single independent equation: Checking the options, for x=11, y=2, 2(11) = 11(2) 22 = 22. Thus, Option 3 satisfies the condition. Whenever you need a high power of a 2x 2 matrix, check its trace and determinant. If |A-lambda I|=0 yields lambda=1,1, then A can be written as I+N where N is nilpotent. This bypasses lengthy matrix multiplication. Class 12 Maths: Matrices and Determinants

For the matrices A=bmatrix3 & -4 1 & -1bmatrix and B=bmatrix-29 & 49 -13 & 18bmatrix, if (A15+B)bmatrixx ybmatrix=bmatrix0 0bmatrix, then among the following which one is true?

(provided N2 = 0, meaning N is nilpotent of order 2). Evaluate powers of matrix A to detect a pattern. Let N = bmatrix2 & -4 1 & -2bmatrix. Check N2: Since N2 = 0, An = I + nN. Given (A15 + B)bmatrixx ybmatrix = bmatrix0 0bmatrix: This yields a single independent equation: Checking the options, for x=11, y=2, 2(11) = 11(2) 22 = 22. Thus, Option 3 satisfies the condition. Whenever you need a high power of a 2x 2 matrix, check its trace and determinant. If |A-lambda I|=0 yields lambda=1,1, then A

For the matrices A=bmatrix3 & -4 1 & -1bmatrix and B=bmatrix-29 & 49 -13 & 18bmatrix, if (A15+B)bmatrixx…

Year	2026	2025	2024	Total
Questions	16	27	16	59

Q jee_main_2025_29_jan_morning Properties of Determinants

Let M and m respectively be the maximum and the minimum values of

f (x) = | arrayc c c 1 + ^ 2 x & ^ 2 x & 4 4 x ^ 2 x & 1 + ^ 2 x & 4 4 x ^ 2 x & ^ 2 x & 1 + 4 4 x array |, x in R

Then

M⁴ -m⁴

is equal to :

A. 1280
B. 1295
C. 1040
D. 1215

Solution

Related Formula

² x + ² x = 1

-1 \leq 4x \leq 1

Core Logic

Apply the row operations $R₂ \to R₂ - R₁$ $R_2 \to R_2 - R_1$ and $R₃ \to R₃ - R₁$ $R_3 \to R_3 - R_1$ to simplify the determinant:

f(x) = | arrayc c c 1 + ^ 2 x & ^ 2 x & 4 4 x -1 & 1 & 0 -1 & 0 & 1 array |

Step 1: Expand the Determinant

Expanding along the first row:

f(x) = (1 + ² x)(1 - 0) - ² x(-1 - 0) + 4 4x(0 - (-1))

f(x) = 1 + ² x + ² x + 4 4x

Since $² x + ² x = 1$ $\sin^2 x + \cos^2 x = 1$, we get:

f(x) = 2 + 4 4x

Step 2: Find Maximum and Minimum Values

The range of $4x$ $\sin 4x$ is $[-1, 1]$ $[-1, 1]$.

M = 2 + 4(1) = 6

m = 2 + 4(-1) = -2

Step 3: Calculate $M⁴ - m⁴$ $M^4 - m^4$

M⁴ - m⁴ = 6⁴ - (-2)⁴ = 1296 - 16 = 1280

Pattern Recognition

Look for repeated structures or cyclic additions in rows. Subtracting rows quickly creates zeros, reducing complex trigonometric matrices into elementary algebraic expressions.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants Class 11 Mathematics: Trigonometric Functions

Q jee_main_2025_29_jan_morning Cofactors and Determinant Value

Let

A = [aᵢⱼ] = bmatrix ₅ 128 & ₄ 5 ₅ 8 & ₄ 25 bmatrix

. If

Aᵢⱼ

is the cofactor of

aᵢⱼ

,

Cᵢⱼ = Σk=1² aik Ajk

,

1 \leq i, j \leq 2

, and

C = [Cᵢⱼ]

, then

8|C|

is equal to:

A. 262
B. 288
C. 242
D. 222

Solution

Related Formula

Σk aik Ajk = δᵢⱼ |A| C = bmatrix |A| & 0 0 & |A| bmatrix |C| = |A|²

Core Logic

Evaluate the determinant of matrix $A$ $A$:

|A| = ( ₅ 128)( ₄ 25) - ( ₄ 5)( ₅ 8)

Using change of base rules:

|A| = (7 ₅ 2)(2 ₄ 5) - ((1)/(2) ₂ 5)(3 ₅ 2)

|A| = 14( ₅ 2 \cdot (1)/(2) ₂ 5) - (3)/(2) = 7 - 1.5 = 5.5 = (11)/(2)

Step 1: Compute |C| and evaluate response target

Since matrix properties dictate $|C| = |A|²$ $|C| = |A|^2$:

|C| = ((11)/(2))² = (121)/(4)

Evaluate targeted multiplier:

8|C| = 8 \times (121)/(4) = 2 \times 121 = 242

Pattern Recognition

Recognize the core cofactor theorem identity instantly: multiplying rows by cofactors of other rows creates zero elements, yielding basic diagonal scalar structures matching matrix attributes.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q jee_main_2025_29_jan_morning Matrix Powers

Let

S = | m in Z : Am² + A^m = 3I - A⁻⁶ |

, where

A = bmatrix 2 & -1 1 & 0 bmatrix

. Then

n(S)

is equal to

Numerical Answer. Answer: 2

Solution

Related Formula

Inductive formulation for exponent powers of a pattern matrix

Core Logic

Evaluate lower power forms of $A$ $A$ to establish inductive sequence patterns:

A = bmatrix 2 & -1 1 & 0 bmatrix, A² = bmatrix 3 & -2 2 & -1 bmatrix, A³ = bmatrix 4 & -3 3 & -2 bmatrix

This cleanly establishes general power state rule configuration expression:

A^m = bmatrix m+1 & -m m & -m+1 bmatrix

Step 1: Setup Matrix Power Equation

Using the pattern, find expressions for terms:

A⁶ = bmatrix 7 & -6 6 & -5 bmatrix, A⁻⁶ = (A⁶)⁻¹ = bmatrix -5 & 6 -6 & 7 bmatrix

Substitute into the targeted equation block:

Am² + A^m = 3 bmatrix 1 & 0 0 & 1 bmatrix - bmatrix -5 & 6 -6 & 7 bmatrix = bmatrix 8 & -6 6 & -4 bmatrix

Step 2: Equate Corresponding Elements

From the bottom-left entry $[2,1]$ $[2,1]$:

m² + m = 6 m² + m - 6 = 0

(m+3)(m-2) = 0 m = -3, 2

Both integer states satisfy all matrix components seamlessly. Therefore, the number of elements $n(S) = 2$ $n(S) = 2$.

Pattern Recognition

Always calculate first 2–3 matrix powers to spot linear arithmetic trends across specific cell values ( $m+1, -m$ $m+1, -m$), avoiding tedious full Cayley-Hamilton character expansions.

Chapter Mix

Class 12 Mathematics: Matrices

Q3 jee_main_2024_01_february_morning Properties of Determinants

If

A= bmatrix\sqrt(2) & 1 -1 & \sqrt(2) bmatrix

,

B= bmatrix1 & 0 1 & 1 bmatrix

,

C=ABAT

and

X=ATC²A

, then

X

is equal to:

A. $243$ $243$
B. $729$ $729$
C. $27$ $27$
D. $891$ $891$

Solution

Related Formula

Properties of Determinants:

$(AB) = (A) \cdot (B)$ $\det(AB) = \det(A) \cdot \det(B)$
$(A^T) = (A)$ $\det(A^T) = \det(A)$
$(Aⁿ) = ( (A))ⁿ$ $\det(A^n) = (\det(A))^n$

Core Logic

First, find the determinant of matrix $A$ $A$ and matrix $B$ $B$:

(A) = |A| = vmatrix\sqrt(2) & 1 -1 & \sqrt(2) vmatrix = (\sqrt(2))(\sqrt(2)) - (1)(-1) = 2 + 1 = 3

(B) = |B| = vmatrix1 & 0 1 & 1 vmatrix = (1)(1) - (0)(1) = 1

Step 1: Calculate determinant of C

Given $C = ABAT$ $C = ABA^{T}$, we calculate its determinant:

(C) = (ABAT) = (A) \cdot (B) \cdot (AT)

Using $(A^T) = (A) = 3$ $\det(A^T) = \det(A) = 3$:

(C) = 3 \cdot 1 \cdot 3 = 9

Step 2: Calculate determinant of X

Given $X = ATC²A$ $X = A^{T}C^{2}A$, we determine $|X|$ $|X|$:

(X) = |ATC²A| = |AT| \cdot |C²| \cdot |A|

(X) = |A| \cdot |C|² \cdot |A| = |A|² \cdot |C|²

Substitute the values $|A| = 3$ $|A| = 3$ and $|C| = 9$ $|C| = 9$:

(X) = (3)² \cdot (9)² = 9 \cdot 81 = 729

Pattern Recognition

Sees: Transpose and multiplication operations nested inside a determinant statement. Shortcut: Never explicitly compute product matrices like $ABA^T$ $ABA^T$ or $A^TC²A$ $A^TC^2A$. Apply the distributive identity of determinants $|XYZ| = |X||Y||Z|$ $|XYZ| = |X||Y||Z|$ entirely to work strictly with scalar multiplication.

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Q12 jee_main_2024_01_february_morning System of Linear Equations

If the system of equations

2x+3y-z=5

x+α y+3z=-4

3x-y+β z=7

has infinitely many solutions, then

13 α β

is equal to:

A. $1110$ $1110$
B. $1120$ $1120$
C. $1210$ $1210$
D. $1220$ $1220$

Solution

Related Formula

For a system of linear equations representing planes to have infinitely many solutions, the planes must belong to a single family sharing a common line of intersection:

P₁ = k₁ P₂ + k₂ P₃

Core Logic

Let the planes be defined as:

$P₁: 2x + 3y - z - 5 = 0$ $P_1: 2x + 3y - z - 5 = 0$
$P₂: x + α y + 3z + 4 = 0$ $P_2: x + \alpha y + 3z + 4 = 0$
$P₃: 3x - y + β z - 7 = 0$ $P_3: 3x - y + \beta z - 7 = 0$

Expressing $P₁$ $P_1$ as a linear combination of $P₂$ $P_2$ and $P₃$ $P_3$:

2x + 3y - z - 5 = k₁(x + α y + 3z + 4) + k₂(3x - y + β z - 7)

Step 1: Evaluate parameters k1 and k2

Comparing the coefficients of $x$ $x$ and the constant terms on both sides:

For $x$ $x$:

k₁ + 3k₂ = 2 (1)

For the constants:

4k₁ - 7k₂ = -5 (2)

Multiplying equation (1) by 4 gives $4k₁ + 12k₂ = 8$ $4k_1 + 12k_2 = 8$. Subtracting equation (2) from this result:

(4k₁ + 12k₂) - (4k₁ - 7k₂) = 8 - (-5)

19k₂ = 13 k₂ = (13)/(19)

Substituting $k₂$ $k_2$ back into equation (1):

k₁ + 3((13)/(19)) = 2 k₁ = 2 - (39)/(19) = -(1)/(19)

Step 2: Calculate alpha, beta and final product

Comparing coefficients for $y$ $y$ and $z$ $z$:

For $y$ $y$:

k₁α - k₂ = 3 -(1)/(19)α - (13)/(19) = 3

-α - 13 = 57 α = -70

For $z$ $z$:

3k₁ + k₂β = -1 3(-(1)/(19)) + (13)/(19)β = -1

-3 + 13β = -19 13β = -16 β = -(16)/(13)

Now, compute $13 α β$ $13 \alpha \beta$:

13 α β = 13 \times (-70) \times (-(16)/(13)) = 70 \times 16 = 1120

Pattern Recognition

Sees: Infinite solution framework for three linear planes. Shortcut: Using the family of planes equation is significantly less prone to fractional algebraic mistakes compared to establishing Cramer's rule determinants ( $D = Dₓ = Dy = Dz = 0$ $D = D_x = D_y = D_z = 0$).

Chapter Mix

Class 12 Mathematics: Matrices and Determinants

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute A^15

Step 2: Evaluate Matrix Operation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Matrices and Determinants Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Expand the Determinant

Step 2: Find Maximum and Minimum Values

Step 3: Calculate M⁴ - m⁴$M^4 - m^4$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute |C| and evaluate response target

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setup Matrix Power Equation

Step 2: Equate Corresponding Elements

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate determinant of C

Step 2: Calculate determinant of X

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluate parameters k1 and k2

Step 2: Calculate alpha, beta and final product

Pattern Recognition

Chapter Mix

More Matrices and Determinants Questions — jee_main_2026_21_jan_evening

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Step 3: Calculate $M⁴ - m⁴$ $M^4 - m^4$