Determinants – Problems
Example 1: Prove that the value of the determinant \(\left| \begin{matrix} -7 & 5+3i & \frac{2}{3}-4i \\ 5-3i & 8 & 4+5i \\ \frac{2}{3}+4i & 4-5i & 9 \\\end{matrix} \right|\) is real.
Solution: Let \(z=\left| \begin{matrix} -7 & 5+3i & \frac{2}{3}-4i \\ 5-3i & 8 & 4+5i \\ \frac{2}{3}+4i & 4-5i & 9 \\\end{matrix} \right|\) … (1)
To prove that this number (z) is real, we have to prove that \(\overline{z}=z\)
Now we know that conjugate of complex number is distributive over all algebraic operations.
Hence, to take conjugate of z in (1), we need not to expand determinant.
To get the conjugate of z, we can take conjugate of each element of determinant. Thus, \(\overline{z}=\left| \begin{matrix} -7 & 5-3i & \frac{2}{3}+4i \\ 5+3i & 8 & 4-5i \\ \frac{2}{3}-4i & 4+5i & 9 \\\end{matrix} \right|\) … (2)
Now interchanging rows into columns (taking transpose) in (2)
We have \(\overline{z}=\left| \begin{matrix} -7 & 5+3i & \frac{2}{3}-4i \\ 5-3i & 8 & 4+5i \\ \frac{2}{3}+4i & 4-5i & 9 \\\end{matrix} \right|\) … (3)
Or \(\overline{z}=z\) [from (1) and (3)] (4)
Hence, z is purely real.
Example 2: If ar = (cos2rπ + isin2rπ)¹/⁹, then prove that \(\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\ {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\\end{matrix} \right|=0\).
Solution: ar = (cos2rπ + isin2rπ)¹/⁹ = \({{e}^{i\frac{2r\pi }{9}}}\).
\(\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\ {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\\end{matrix} \right|\).
\(=\left| \begin{matrix} {{e}^{i\frac{2\pi }{9}}} & {{e}^{i\frac{4\pi }{9}}} & {{e}^{i\frac{6\pi }{9}}} \\ {{e}^{i\frac{8\pi }{9}}} & {{e}^{i\frac{10\pi }{9}}} & {{e}^{i\frac{12\pi }{9}}} \\ {{e}^{i\frac{14\pi }{9}}} & {{e}^{i\frac{16\pi }{9}}} & {{e}^{i\frac{18\pi }{9}}} \\\end{matrix} \right|\).
\(={{e}^{i\frac{6\pi }{9}}}\left| \begin{matrix} {{e}^{i\frac{2\pi }{9}}} & {{e}^{i\frac{4\pi }{9}}} & {{e}^{i\frac{6\pi }{9}}} \\ {{e}^{i\frac{8\pi }{9}}} & {{e}^{i\frac{10\pi }{9}}} & {{e}^{i\frac{12\pi }{9}}} \\ {{e}^{i\frac{14\pi }{9}}} & {{e}^{i\frac{16\pi }{9}}} & {{e}^{i\frac{18\pi }{9}}} \\\end{matrix} \right|\) [taking \({{e}^{i\frac{6\pi }{9}}}\) common from R₂
= 0 [R₁ and R₂ are identical].
Example 3: Without expanding the determinants, prove that \(\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|+\left| \begin{matrix} 113 & 116 & 104 \\ 108 & 106 & 111 \\ 115 & 114 & 103 \\\end{matrix} \right|=0\).
Solution: \(\underset{{{D}_{1}}}{\mathop{D=\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|}}\,+\underset{{{D}_{2}}}{\mathop{\left| \begin{matrix} 113 & 116 & 104 \\ 108 & 106 & 111 \\ 115 & 114 & 103 \\\end{matrix} \right|}}\,\).
In D₂, interchanging C₁ and C₃,
\(\underset{{{D}_{1}}}{\mathop{D=\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|}}\,-\underset{{{D}_{2}}}{\mathop{\left| \begin{matrix} 104 & 116 & 113 \\ 111 & 106 & 108 \\ 103 & 114 & 115 \\\end{matrix} \right|}}\,\).
In D₂, interchanging C₂ and C₃,
\(\underset{{{D}_{1}}}{\mathop{D=\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix}\right|}}\,+\underset{{{D}_{2}}}{\mathop{\left| \begin{matrix} 104 & 113 & 116 \\ 111 & 108 & 106 \\ 103 & 115 & 114 \\\end{matrix} \right|}}\,\).
In D₂, interchanging R₁ and R₃,
\(\underset{{{D}_{1}}}{\mathop{D=\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|}}\,-\underset{{{D}_{2}}}{\mathop{\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|=0}}\,\).