Question

Let $A$ and $B$ be two $3\times 3$ invertible matrices such that $A$ is an idempotent matrix and
$det\left(adjB\right)=det(A{)}^{12}+det(A-A^T{)}^{13}+det\left(adj\right(A)-I{)}^{14}+L,$
where $L=\underset{x\to 0}{lim}\frac{\underset{0}{\overset{{\sin }^{2}x}{\int }}\ln (1+t^3)dt}{\underset{0}{\overset{x}{\int }}(e^t-1-t)(\tan t-t)(1-\cos t)dt}.$
The absolute value of $det\left(B\right)$ is

Answer 1

$det\left(adjB\right)=det(A{)}^{12}+det(A-A^T{)}^{13}+det\left(adj\right(A)-I{)}^{14}+L$
As $A$ is an idempotent matrix and invertible also, so
$A^2=A\Rightarrow A^{-1}{A}^2=A^{-1}A$
$\therefore A=I$ where $I$ is an indentity matrix of order $3\times 3$
Now, $det(A{)}^{12}=1$
$det(A-A^T{)}^{13}=0det(adjA-I{)}^{14}=0$

Now,
$L=\underset{x\to 0}{lim}\frac{\underset{0}{\overset{{\sin }^{2}x}{\int }}\ln (1+t^3)dt}{\underset{0}{\overset{x}{\int }}(e^t-1-t)(\tan t-t)(1-\cos t)dt}$
Using L'Hospital's Rule, we get
$L=\underset{x\to 0}{lim}\frac{\ln (1+{\sin }^{6}x)2\sin x\cos x}{(e^x-1-x)(\tan x-x)(1-\cos x)}\Rightarrow L=\underset{x\to 0}{lim}\frac{2x{\sin }^{6}x}{(e^x-1-x)(\tan x-x)\frac{(1-\cos x)}{x^2}{x}^2}\Rightarrow L=\underset{x\to 0}{lim}\frac{4x^5}{(e^x-1-x)(\tan x-x)}(\because \underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=\frac{1}{2})$

We know that
$L_1=\underset{x\to 0}{lim}\frac{\tan x-x}{x^3}\Rightarrow L_1=\underset{x\to 0}{lim}\frac{{\sec }^{2}x-1}{3x^2}\Rightarrow L_1=\underset{x\to 0}{lim}\frac{{\tan }^{2}x}{3x^2}=\frac{1}{3}$
Therefore, $L=\underset{x\to 0}{lim}\frac{12x^2}{e^x-1-x}$
$\Rightarrow L=\underset{x\to 0}{lim}\frac{24x}{e^x-1}=24$
So, $det\left(adjB\right)=1+0+0+24$
$\Rightarrow |B{|}^2=25\Rightarrow |B|=\pm 5$
Hence, the absolute value of $det\left(B\right)=5$