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If $A=\left[\begin{array}{ccc}1 & 1 & a+1 \\ 1 & a+1 & 1 \\ a+1 & 1 & 1\end{array}\right]$ is not an invertible matrix, then the sum of all the values of $a$ is
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The correct answer is:
$-3$
Given, $A=\left[\begin{array}{ccc}1 & 1 & a+1 \\ 1 & a+1 & 1 \\ a+1 & 1 & 1\end{array}\right]$
$\because A$ is non-invertible matrix.
$\begin{aligned} & \Rightarrow|A|=0 \\ & \Rightarrow 1(a+1-1)-(1-a-1)+(a+1)\left(1-(a+1)^2\right)=0 \\ & \Rightarrow a+a+a+1-(a+1)^3=0 \\ & \Rightarrow \quad 3 a+1-a^3-1-3 a^2-3 a=0 \\ & \Rightarrow-a^3-3 a^2=0 \Rightarrow-a^2(a+3)=0 \\ & \Rightarrow a=0,-3\end{aligned}$
Sum of all values of $a=0+(-3)=-3$
$\because A$ is non-invertible matrix.
$\begin{aligned} & \Rightarrow|A|=0 \\ & \Rightarrow 1(a+1-1)-(1-a-1)+(a+1)\left(1-(a+1)^2\right)=0 \\ & \Rightarrow a+a+a+1-(a+1)^3=0 \\ & \Rightarrow \quad 3 a+1-a^3-1-3 a^2-3 a=0 \\ & \Rightarrow-a^3-3 a^2=0 \Rightarrow-a^2(a+3)=0 \\ & \Rightarrow a=0,-3\end{aligned}$
Sum of all values of $a=0+(-3)=-3$
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