3.8 (h) Unitary Matrix :

1.1 Type of Matrices
1.2 Trace of a Matrix:
1.3 Transpose of a Matrix
1.4 Conjugate of a Matrix

2.1 (a) Addition and subtraction of Matrices:
2.2 (b) Multiplication of Matrices:

3.1 (a) Symmetric Matrix:
3.2 (b) Skew Symmetric Matrix:
3.3 (c) Hermitian matrix:
3.4 (d) Skew Hermitian Matrix:
3.5 (e) Singular and Non-singular Matrices:
3.6 (f) Orthogonal Matrix :
3.7 (g) Idempotent Matrix :
3.8 (h) Unitary Matrix :
3.9 (i) Involuntary Matrix:
3.10 (j) Nilpotent Matrix:

4.1 Minors & Cofactors
4.2 Properties of Determinant

5.1 Elementary row Transformations

6.1 Properties of Inverse of a Matrix
6.2 Elementary row Transformations

8.1 Echelon form of a Matrix
8.2 Rank of a Matrix

A square matrix $A$ is said to be unitary if ${A^\theta }A = I$ , where $I$ is an identity matrix & ${A^\theta }$ is the transposed conjugate of $A$.

Prove that the matrix $\frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right]$ is unitary.

Let $A = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right]$

${A^T} = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 - i} \\ {1 + i}&{ - 1} \end{array}} \right]$

$\overline {\left( {{A^T}} \right)} = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right]$

${A^\theta } = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right]$

$A{A^\theta } = \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right] \times \frac{1}{{\sqrt 3 }}\left[ {\begin{array}{c} 1&{1 + i} \\ {1 - i}&{ - 1} \end{array}} \right]$

$ = \frac{1}{3}\left[ {\begin{array}{c} 3&0 \\ 0&3 \end{array}} \right] = \left[ {\begin{array}{c} 1&0 \\ 0&1 \end{array}} \right] = I$

Hence $A$ is unitary matrix.