Question

If the matrix B given as $B=\left[ \begin{matrix}
   5 & 2\alpha & 1 \\
   0 & 2 & 1 \\
   \alpha & 3 & -1 \\
\end{matrix} \right]$ is the inverse of a $3\times 3$ matrix A, then the sum of all values of $\alpha $ for which $\det \left( A \right)+1=0$, is?
$\begin{align}
  & \left( A \right)0 \\
 & \left( B \right)2 \\
 & \left( C \right)1 \\
 & \left( D \right)-1 \\
\end{align}$

Answer 1

Hint: We solve this problem finding the relation between the determinant of A and determinant of B using the formula $\left| AB \right|=\left| A \right|\left| B \right|$. Then we find the determinant of B and substitute in the relation obtained to find the determinant of A. Then we substitute the determinant of A value in $\det \left( A \right)+1=0$ and find the values of $\alpha $ and then we find the sum of those values.

Complete step-by-step answer:
Two matrices A and B are said to be inverse of each other, if $AB=BA=I$

We are given that matrix B is $B=\left[ \begin{matrix}
   5 & 2\alpha & 1 \\
   0 & 2 & 1 \\
   \alpha & 3 & -1 \\
\end{matrix} \right]$ and it is inverse of matrix A.
As, A and B are inverses of each other we can write it as,
$\Rightarrow AB=I...........\left( 1 \right)$
Let us consider the property of determinants. Determinant of product of matrices is equal to the product of determinant of matrices.
$\left| AB \right|=\left| A \right|\left| B \right|$
Now let us apply determinant to the above equation (1). Then we get,
$\begin{align}
  & \Rightarrow \left| AB \right|=\left| I \right| \\
 & \Rightarrow \left| A \right|\left| B \right|=1 \\
 & \Rightarrow \left| A \right|=\dfrac{1}{\left| B \right|} \\
\end{align}$
So, we get that $\det \left( A \right)=\dfrac{1}{\det \left( B \right)}$.
Now, let us find the determinant of matrix B.
\[\begin{align}
  & \left| B \right|=\left| \begin{matrix}
   5 & 2\alpha & 1 \\
   0 & 2 & 1 \\
   \alpha & 3 & -1 \\
\end{matrix} \right| \\
 & \left| B \right|=5\left| \begin{matrix}
   2 & 1 \\
   3 & -1 \\
\end{matrix} \right|-2\alpha \left| \begin{matrix}
   0 & 1 \\
   \alpha & -1 \\
\end{matrix} \right|+\left| \begin{matrix}
   0 & 2 \\
   \alpha & 3 \\
\end{matrix} \right| \\
 & \left| B \right|=5\left( -2-3 \right)-2\alpha \left( 0-\alpha \right)+\left( 0-2\alpha \right) \\
 & \left| B \right|=5\left( -5 \right)-2\alpha \left( -\alpha \right)+\left( -2\alpha \right) \\
 & \left| B \right|=-25+2{{\alpha }^{2}}-2\alpha \\
\end{align}\]
As $\det \left( A \right)=\dfrac{1}{\det \left( B \right)}$, let us now calculate the value of determinant of A. Then we get
$\det \left( A \right)=\dfrac{1}{\det \left( B \right)}=\dfrac{1}{2{{\alpha }^{2}}-2\alpha -25}$
Now let us consider $\det \left( A \right)+1=0$.
$\begin{align}
  & \Rightarrow \dfrac{1}{2{{\alpha }^{2}}-2\alpha -25}+1=0 \\
 & \Rightarrow \dfrac{1+2{{\alpha }^{2}}-2\alpha -25}{2{{\alpha }^{2}}-2\alpha -25}=0 \\
 & \Rightarrow 2{{\alpha }^{2}}-2\alpha -24=0 \\
 & \Rightarrow 2\left( {{\alpha }^{2}}-\alpha -12 \right)=0 \\
 & \Rightarrow {{\alpha }^{2}}-\alpha -12=0 \\
 & \Rightarrow \left( \alpha -4 \right)\left( \alpha +3 \right)=0 \\
 & \Rightarrow \alpha =4,-3 \\
\end{align}$
So, the possible values of $\alpha $ for which $\det \left( A \right)+1=0$ are -3 and 4.
So, sum of the possible values of $\alpha $ for which $\det \left( A \right)+1=0$ is
$\Rightarrow 4+\left( -3 \right)=1$
Hence the answer is 1.
Therefore, the answer is Option C.

Note:There is a possibility of making a mistake while solving this problem by taking the determinant of matrix A equal to the determinant of matrix B when they are inverse to each other. After obtaining the quadratic equation, we can also use the quadratic formula to find the possible values of $\alpha $ .