If A = begin{bmatrix} 1 & 1 & a+1 1 & a+1 & | vedclass

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(A) Given,$A = \begin{bmatrix} 1 & 1 & a+1 \ 1 & a+1 & 1 \ a+1 & 1 & 1 \end{bmatrix}$.Since $A$ is a non-invertible matrix,its determinant must be z

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$-3$

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(A) Given,$A = \begin{bmatrix} 1 & 1 & a+1 \ 1 & a+1 & 1 \ a+1 & 1 & 1 \end{bmatrix}$.Since $A$ is a non-invertible matrix,its determinant must be zero,i.e.,$|A| = 0$.Calculating the determinant along the first row:$|A| = 1((a+1)(1) - 1(1)) - 1(1(1) - 1(a+1)) + (a+1)(1(1) - (a+1)(a+1)) = 0$$|A| = 1(a+1-1) - 1(1-a-1) + (a+1)(1-(a+1)^2) = 0$$|A| = a + a + (a+1)(1 - (a^2 + 2a + 1)) = 0$$|A| = 2a + (a+1)(-a^2 - 2a) = 0$$|A| = 2a - a^3 - 2a^2 - a^2 - 2a = 0$$-a^3 - 3a^2 = 0$$-a^2(a+3) = 0$Thus,the values of $a$ are $a = 0$ and $a = -3$.The sum of all values of $a$ is $0 + (-3) = -3$.

If $A = \begin{bmatrix} 1 & 1 & a+1 \\ 1 & a+1 & 1 \\ a+1 & 1 & 1 \end{bmatrix}$ is not an invertible matrix,then the sum of all the values of $a$ is

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