
If we have a determinant as \[Z=\left| \begin{matrix}
1 & 1+2i & -5i \\
1-2i & -3 & 5+3i \\
5i & 5-3i & 7 \\
\end{matrix} \right|\] then \[(i=-\sqrt{-1})\]
A) \[z\] is purely real
B) \[z\]is purely imaginary
C) \[z+\overline{z}=0\]
D) \[z-\overline{z}=0\] is purely imaginary
Answer
480.6k+ views
Hint: \[i\] stands for iota. It is also called the imaginary part. A complex number has the real part and the imaginary part. The complex numbers are the field of numbers of the form \[x+iy\] , where and \[y\] are real numbers and \[i\] is the imaginary unit equal to the square root of \[-1\],\[\sqrt{-1}\] . When a single letter \[z=x+iy\] is used to denote a complex number, it is also called an "affix." In component notation, \[z=x+iy\] can be written\[(x,y)\]. The field of complex numbers includes the field of real numbers as a subfield. An \[m\times n\] complex matrix is a rectangular array of complex numbers arranged in \[m\] rows and \[n\] columns. Addition and scalar multiplication of complex matrices can be performed. We can represent the complex number \[z=x+iy\] as the matrix \[\left[ \begin{matrix}
x & -y \\
y & x \\
\end{matrix} \right]\] . The reciprocal of a complex number is given as the inverse of its matrix representation; we can see that complex division can be given as multiplying the matrix representation of the numerator by the inverse matrix of the denominator. If \[x=0\], then \[z=0+iy\] is purely imaginary. If \[y=0\], then \[z=x+0i\] is purely real.
Complete step-by-step solution:
According to the question,
\[Z=\left| \begin{matrix}
1 & 1+2i & -5i \\
1-2i & -3 & 5+3i \\
5i & 5-3i & 7 \\
\end{matrix} \right|\]
Using matrix modulus method we get
\[=1(-21-(25+9))-(1+2i)(7-14i-(25i-15))-5i[(-1-13i)+15i]\]
Simplifying we get
\[=-55-100-5i+5i+10\]
\[=-145\]
Therefore, \[z\] is purely real.
Hence, option \[(1)\] is the correct answer.
Note: To solve this type of questions all the properties of matrices and the complex matrices have to be known properly. Two complex numbers are equal if and only if their real and imaginary numbers are equal. A matrix whose elements contain complex numbers is called a complex matrix. The sum of two conjugate complex numbers is always real.
x & -y \\
y & x \\
\end{matrix} \right]\] . The reciprocal of a complex number is given as the inverse of its matrix representation; we can see that complex division can be given as multiplying the matrix representation of the numerator by the inverse matrix of the denominator. If \[x=0\], then \[z=0+iy\] is purely imaginary. If \[y=0\], then \[z=x+0i\] is purely real.
Complete step-by-step solution:
According to the question,
\[Z=\left| \begin{matrix}
1 & 1+2i & -5i \\
1-2i & -3 & 5+3i \\
5i & 5-3i & 7 \\
\end{matrix} \right|\]
Using matrix modulus method we get
\[=1(-21-(25+9))-(1+2i)(7-14i-(25i-15))-5i[(-1-13i)+15i]\]
Simplifying we get
\[=-55-100-5i+5i+10\]
\[=-145\]
Therefore, \[z\] is purely real.
Hence, option \[(1)\] is the correct answer.
Note: To solve this type of questions all the properties of matrices and the complex matrices have to be known properly. Two complex numbers are equal if and only if their real and imaginary numbers are equal. A matrix whose elements contain complex numbers is called a complex matrix. The sum of two conjugate complex numbers is always real.
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