Let a R and A be a matrix of order 3 3 such that (A)=-4 and A+I=[[|gathered 1 a 1 2 1 0 a …

Question

Let $a\in R$ and $A$ be a matrix of order $3	imes 3$ such that $\det (A)=-4$ and $A+I=[[|\begin{gathered} 1 \quad  a \quad  1 \ 2 \quad  1 \quad  0 \ a \quad  1 \quad  2 \end{gathered}]]|$, where $I$ is the identity matrix of order $3	imes 3$. If $\det ((a+1)adj((a-1)A))$ is ${2}^{m}{3}^{n},m$, $n\in {0,1,2,\ldots ,20}$, then $m+n$ is equal to :

Accepted Answer

To solve for the value of $ m+n $, we first establish the matrix $ A $ and determine the value of $ a $. The condition given is:

$ A + I = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix} $

Since $ I $ is the identity matrix of size $ 3 	imes 3 $, we have:

$ A = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix} - I = \begin{bmatrix} 0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1 \end{bmatrix} $

We know $\operatorname{det}(A) = -4$. Calculating the determinant:

$ \det(A) = 0(0 	imes 1 - 0 	imes 1) - a(2 	imes 1 - 0 	imes a) + 1(2 	imes 1 - 0 	imes 1) = -2a + 2 $

Setting $-2a + 2 = -4$, we solve for $ a $:

$ -2a + 2 = -4 $

$ -2a = -6 $

$ a = 3 $

With $ a = 3 $, the problem is to find $\operatorname{det}((a+1) \operatorname{adj}((a-1) A))$. With $ a = 3 $:

Calculate $(a+1) \operatorname{adj}((a-1) A)$:

$    = 4 \operatorname{adj}(3A)     $

Calculate the determinant:

$    \det(4 \operatorname{adj}(3A)) = 4^3 	imes \det(\operatorname{adj}(3A))    $

Using the property $\operatorname{det}(\operatorname{adj}(A)) = \det(A)^{n-1}$ for a $ n 	imes n $ matrix:

$    \operatorname{adj}(3A) = (3A)^{2-1} = (3 	imes \det(A))^2    $

$    = (3^3 	imes (-4))^2    $

Therefore, calculate $ \det(3A)^2 $:

$    \det((3A)^2) = (3^3 	imes (-4))^2 = 3^6 	imes 4^2    $

Substitute back:

$    4^3 	imes \det((3A)^2) = 4^3 	imes 3^6 	imes 4^2 = 2^6 	imes 3^6 	imes 2^4    $

Simplifying gives:

$    = 2^{10} 	imes 3^6    $

Finally, we have powers $ m = 10 $ and $ n = 6 $. Therefore:

$ m+n = 10 + 6 = 16 $

Solutions · 1