Question

Let a,b,c such that $b\left( a+c \right)\ne 0$. If \[\left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+\left| \begin{matrix}
   a+1 & b+1 & c-1 \\
   a-1 & c-1 & c+1 \\
   {{\left( -1 \right)}^{n+2}}a & {{\left( -1 \right)}^{n+1}}b & {{\left( -1 \right)}^{n}}c \\
\end{matrix} \right|=0\], then then the value of $n$ is\[\]
A. Any integer\[\]
B. zero\[\]
C. Any even integer\[\]
D. Any odd integer\[\]

Answer 1

Hint: We first take ${{\left( -1 \right)}^{n}}$ common from the third row of the second determinant and take it outside of the determinant as multiple. We take the transpose of the second determinant and exchange columns two times such that the second determinant can be expressed in terms of the first determinant. We take the first determinant common and solve the equation for $n$.\[\]

Complete step-by-step solution:
We know that the transpose of a matrix $A$ with $m$ rows and $n$ column is denoted as ${{A}^{T}}$ and can be obtained by \[\]
1. Reflecting $A$ over its main diagonal (which runs from top-left to bottom-right)\[\]
2. Writing the rows of $A$ as the columns of ${{A}^{T}}$\[\]
3. Writing the columns of $A$ as the rows of ${{A}^{T}}$\[\]
We can only find the determinant of the matrix if the matrix is a square matrix where the determinant value of the matrix $A$ and determinant value of the transpose are equal which means
\[\det \left( A \right)=\det \left( {{A}^{T}} \right)\]
We know that we can interchange rows $n$ a number of times if we multiply ${{\left( -1 \right)}^{n}}$ to the determinant. We also know that if we multiply a multiple $k$ with any row or column then the determinant value changes by $k$ times which means if $k$ is a factor every element in a row or column we can take $k$ outside the determinant as a factor.
We are given in the question an equation of determinants
\[\left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+\left| \begin{matrix}
   a+1 & b+1 & c-1 \\
   a-1 & c-1 & c+1 \\
   {{\left( -1 \right)}^{n+2}}a & {{\left( -1 \right)}^{n+1}}b & {{\left( -1 \right)}^{n}}c \\
\end{matrix} \right|=0\]
Let us take ${{\left( -1 \right)}^{n}}$ common from the third row of the second determinant and take outside of determinant as multiple to have
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+\left| \begin{matrix}
   a+1 & b+1 & c-1 \\
   a-1 & c-1 & c+1 \\
   {{\left( -1 \right)}^{n}}\cdot {{\left( -1 \right)}^{2}}a & {{\left( -1 \right)}^{n}}\cdot {{\left( -1 \right)}^{1}}b & {{\left( -1 \right)}^{n}}c \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n}}\left| \begin{matrix}
   a+1 & b+1 & c-1 \\
   a-1 & c-1 & c+1 \\
   a & -b & c \\
\end{matrix} \right|=0 \\
\end{align}\]
Let us take transpose of second determinant to have,
\[\Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n}}\left| \begin{matrix}
   a+1 & a-1 & a \\
   b+1 & b-1 & -b \\
   c-1 & c-1 & c \\
\end{matrix} \right|=0\]
We exchange third column with second column $\left( {{C}_{2}}\leftrightarrow {{C}_{3}} \right)$ in second determinant to have to have,
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n}}\cdot {{\left( -1 \right)}^{1}}\left| \begin{matrix}
   a+1 & a & a-1 \\
   b+1 & -b & b-1 \\
   c-1 & c & c-1 \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n+1}}\left| \begin{matrix}
   a+1 & a & a-1 \\
   b+1 & -b & b-1 \\
   c-1 & c & c-1 \\
\end{matrix} \right|=0 \\
\end{align}\]
We exchange first column with second column $\left( {{C}_{1}}\leftrightarrow {{C}_{2}} \right)$ in second determinant to have to have,
\[\begin{align}
  & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n+1}}\cdot {{\left( -1 \right)}^{1}}\left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c-1 \\
\end{matrix} \right|=0 \\
 & \Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|+{{\left( -1 \right)}^{n+2}}\left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c-1 \\
\end{matrix} \right|=0 \\
\end{align}\]
We take the first determinant common and have
\[\Rightarrow \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right|\left( 1+{{\left( -1 \right)}^{n+2}} \right)=0\]
We observe that the equation is valid when $1+{{\left( -1 \right)}^{n+2}}=0$ or the value of the first determinant is zero. Let us find the value of the first determinant expanding by the first column. We have,
\[\begin{align}
  & \left| \begin{matrix}
   a & a+1 & a-1 \\
   -b & b+1 & b-1 \\
   c & c-1 & c+1 \\
\end{matrix} \right| \\
&=a\left[ bc+b+c+1-\left( bc-b-c+1 \right) \right]-\left( -b \right)\left[ ac+a+c+1-\left( ac-a-c+1 \right) \right]+c\left[ ab-a+b-1-\left( ab+a-b-1 \right) \right] \\
 & =2ab+2ac+2ab+2bc-2ac+2bc \\
 & =4b\left( a+c \right) \\
\end{align}\]
The determinant value $4b\left( a+c \right)$ cannot be zero since we are given in the question that $b\left( a+c \right)\ne 0$ and hence we conclude$1+{{\left( -1 \right)}^{n+2}}=0$; which is possible only when ${{\left( -1 \right)}^{n+2}}=-1$ which implies $n$ is an odd integer.
So the correct option is C.

Note: We note that if we would have been given the condition $b\left( a+c \right)=0$ then $n$ is any integer to satisfy the equation. If we take two times the transpose we shall get the original matrix ${{\left( {{A}^{T}} \right)}^{T}}=A$ , it is called involution. The determinant of inverse of matrix is the reciprocal of determinant which means$\det \left( {{A}^{-1}} \right)=\dfrac{1}{\det \left( A \right)}$.