If A = bmatrix 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 bmatrix, then the value of, A^2025, is Option 1: 1 Option 2: 126 Option 3: -126 Option 4: 0

Answer: 0. Solution: We need to find the determinant of A raised to the power 2025, written as, A^2025, . For any matrix M and positive integer n:, M^n, =, M, ^n Therefore:, A^2025, =, A, ^2025 The given matrix is: A = bmatrix 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 bmatrix Observe that: - All diagonal elements are 0 - The element at position (1,2) is 1, and at position (2,1) is -1 - The element at position (1,3) is -3, and at position (3,1) is 3 - The element at position (2,3) is 5, and at position (3,2) is -5 Each pair of opposite elements have opposite signs. This is a skew-symmetric matrix where A^T = -A. For any skew-symmetric matrix of odd order (like 3×3), the determinant is always 0. Since matrix A is 3×3 (odd order) and skew-symmetric:, A, = 0 Using the property, A^2025, =, A, ^2025:, A^2025, =, A, ^2025 = 0^2025 = 0 Therefore, the value of, A^2025, = 0.

Algebra

Matrices & Determinants

Medium

core

If $A = \begin{bmatrix} 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}$ , then the value of $|A^{2025}|$ is

126

-126

✅ Correct Option: 4

We need to find the determinant of $A$ raised to the power 2025, written as $|A^{2025}|$ .

For any matrix $M$ and positive integer $n$ :

$|M^n| = |M|^n$

Therefore:

$|A^{2025}| = |A|^{2025}$

The given matrix is:

$A = \begin{bmatrix} 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}$

Observe that:

All diagonal elements are 0
The element at position (1,2) is 1, and at position (2,1) is -1
The element at position (1,3) is -3, and at position (3,1) is 3
The element at position (2,3) is 5, and at position (3,2) is -5

Each pair of opposite elements have opposite signs. This is a skew-symmetric matrix where $A^T = -A$ .

For any skew-symmetric matrix of odd order (like 3×3), the determinant is always 0.

Since matrix $A$ is 3×3 (odd order) and skew-symmetric:

$|A| = 0$

Using the property $|A^{2025}| = |A|^{2025}$ :

$|A^{2025}| = |A|^{2025}$

$= 0^{2025}$

$= 0$

Therefore, the value of $|A^{2025}| = 0$ .

2025: 29 May Shift 2

If matrix $A_p = \begin{bmatrix} p & p+1 \\ p-1 & p \end{bmatrix}_{p \in N}$ (where N is the set of natural numbers), then the value of $|A_1| + |A_2| + |A_3| + \dots + |A_{2025}|$ is:

2025: 15 May Shift 2

If $\begin{bmatrix} 1 & 0 \\ b & 5 \end{bmatrix} + 2\begin{bmatrix} a & 0 \\ 1 & -2 \end{bmatrix} = I$ , where $I$ is a unit matrix of order 2, then the value of $(a - b)$ is:

2026: 18 May Shift 2

If $x,y,z$ are non-zero real numbers such that $\begin{vmatrix} 10+x & 10 & 10 \\ 10 & 10+y & 10 \\ 10 & 10 & 10+z \end{vmatrix}=0$ , then the value of $x^{-1}+y^{-1}+z^{-1}$ is

2026: 19 May Shift 1

If $A = \begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix}$ , then which of the following statements are TRUE?

[where $A_{ij}$ is the cofactor of the (i, j)th element, $a_{ij}$ of matrix A, $1 \gt i, j \le 3$ .]

(A) $a_{11}A_{31} = -24$

(B) $a_{12}A_{32} = -66$

(D) $a_{21}A_{21} = -24$

Choose the correct answer from the options given below:

2022: 4 Aug Shift 1

Match List I with List II: Given that A and B are invertible matrices of size $3 \times 3$

List I	List II
A. $\lvert AB \rvert$	I. $\frac{1}{\lvert A \rvert}$
B. $\lvert \operatorname{Adj} A \rvert$	II. $\lvert A \rvert \lvert B \rvert$
C. $\lvert A^{-1} \rvert$	III. $B^{-1} \cdot A^{-1}$
D. $(AB)^{-1}$	IV. $\lvert A \rvert^2$

Choose the correct answer from the options given below:

2026: 19 May Shift 2

If A and B are square matrices of same order such that, $AB=BA$ , $A^{-1} = A^2$ and $B^{-1} = B^3$ , then the value of $(AB)^{2040}$ , is

If A=[01−3−1053−50]A = \begin{bmatrix} 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}A=​0−13​10−5​−350​​, then the value of ∣A2025∣|A^{2025}|∣A2025∣ is

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If $A = \begin{bmatrix} 0 & 1 & -3 \\ -1 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix}$ , then the value of $|A^{2025}|$ is