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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(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

Similar Questions

If the matrix $\begin{bmatrix} 1 & 3 & \lambda + 2 \\ 2 & 4 & 8 \\ 3 & 5 & 10 \end{bmatrix}$ is singular,then $\lambda = $

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to

If $\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x, y, z$ are all distinct,then $x y z=$

The least value of the product $xyz$ for which the determinant $\left| \begin{array}{ccc} x & 1 & 1 \\ 1 & y & 1 \\ 1 & 1 & z \end{array} \right|$ is non-negative,is

Let $x, y, z > 0$ be the $2^{nd}, 3^{rd}, 4^{th}$ terms of a $G.P.$ respectively,and $\Delta = \begin{vmatrix} x^k & x^{k+1} & x^{k+2} \\ y^k & y^{k+1} & y^{k+2} \\ z^k & z^{k+1} & z^{k+2} \end{vmatrix} = (r-1)^2 \left(1 - \frac{1}{r^2}\right)$,where $r$ is the common ratio. Then $k = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module