Question

What is the determinant of an identity matrix of any order \(n\)?

• A. \(0\)
• B. \(1\)
• C. \(n\)
• D. \(-1\)

Hint:

The determinant of any {identity matrix} is always \(1\), regardless of its size. Also, multiplying any matrix by an identity matrix leaves the matrix unchanged.

Answer 1

The Correct Option is B

Concept: An identity matrix \(I_n\) of order \(n\) is a square matrix in which:

• All diagonal elements are \(1\).
• All non-diagonal elements are \(0\).

The general form of an identity matrix is: \[ I_n = \begin{bmatrix} 1 & 0 & 0 & \dots & 0
0 & 1 & 0 & \dots & 0
0 & 0 & 1 & \dots & 0
\vdots & \vdots & \vdots & \ddots & \vdots
0 & 0 & 0 & \dots & 1 \end{bmatrix} \] A key property of determinants is that the determinant of a triangular matrix (including diagonal matrices) equals the product of its diagonal elements.
Step 1: Identify the diagonal elements.
For an identity matrix \(I_n\), all diagonal elements are equal to \(1\). \[ 1, 1, 1, \dots , 1 \]
Step 2: Apply the determinant rule.
The determinant of a diagonal matrix is the product of its diagonal elements. \[ \det(I_n) = 1 \times 1 \times 1 \times \dots \times 1 \]
Step 3: Compute the result.
Since the product of \(n\) ones is: \[ \det(I_n) = 1 \]
Step 4: Conclusion.
Therefore, the determinant of an identity matrix of any order \(n\) is: \[ 1 \]