# Determinants – Problems

## 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,

$$\underset{{{D}_{1}}}{\mathop{D=\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|}}\,-\underset{{{D}_{2}}}{\mathop{\left| \begin{matrix} 103 & 115 & 114 \\ 111 & 108 & 106 \\ 104 & 113 & 116 \\\end{matrix} \right|=0}}\,$$.