For a matrix $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)$ if $U_{1},U_{2}$ and $U_{3}$ are $3 \times 1$ column matrices satisfying $A U_{1}=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right), A U_{2}=\left(\begin{array}{l}2 \\ 3 \\ 0\end{array}\right), A U_{3}=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right)$ and $U$ is $3 \times 3$ matrix whose columns are $U_{1}, U_{2}$ and $U_{3}$. Then, sum of he elements of $U^{-1}$ is

Options:
A 6
B 0
C 1
D $2 / 3$

The correct answer is:
0

Solution:
Let $U_{i}=\left(\begin{array}{l}a_{i} \\ b_{1} \\ c_{i}\end{array}\right),$ where $i=1,2,3$
$\therefore \quad A U_{1}=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\left(\begin{array}{l}a_{1} \\ b_{1} \\ c_{1}\end{array}\right)$
$\Rightarrow\left(\begin{array}{c}a_{1} \\ 2 a_{1}+b_{1} \\ 3 a_{1}+2 b_{1}+c_{1}\end{array}\right)=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right) \quad\left[\because A U_{1}=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)\right]$
$\therefore \quad a_{1}=1,2 a_{1}+b_{1}=0$
$\Rightarrow \quad 2+b_{1}=0$
$\Rightarrow \quad b_{1}=-2$
and $3 a_{1}+2 b_{1}+c_{1}=0$
$\Rightarrow \quad 3+2(-2)+c_{1}=0$
$\Rightarrow \quad 3-4+c_{1}=0$
$\Rightarrow \quad c_{1}=1$
Similarly, $A U_{2}=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\left(\begin{array}{l}a_{2} \\ b_{2} \\ c_{2}\end{array}\right)$
$\Rightarrow\left(\begin{array}{c}a_{2} \\ 2 a_{2}+b_{2} \\ 3 a_{2}+2 b_{2}+c_{2}\end{array}\right)=\left(\begin{array}{l}2 \\ 3 \\ 0\end{array}\right) \quad\left[\because A U_{2}=\left(\begin{array}{l}2 \\ 3 \\ 0\end{array}\right)\right]$
$\therefore a_{2}=2$ and $2 a_{2}+b_{2}=3$
$\Rightarrow \quad 2 \times 2+b_{2}=3$
$\Rightarrow \quad 4+b_{2}=3$
$\Rightarrow \quad b_{2}=-1$
and $\quad 3 a_{2}+2 b_{2}+c_{2}=0$
$\Rightarrow \quad 3 \times 2+2(-1)+c_{2}=0$
$\Rightarrow \quad 6-2+c_{2}=0$
$\Rightarrow \quad c_{2}=-4$
Now, $A U_{3}=\left(\begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right)\left(\begin{array}{l}a_{3} \\ b_{3} \\ c_{3}\end{array}\right)$
$\Rightarrow\left(\begin{array}{c}a_{3} \\ 2 a_{3}+b_{3} \\ 3 a_{3}+2 b_{3}+c_{3}\end{array}\right)=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right) \quad\left[\because A U_{3}=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right)\right]$
$\therefore \quad a_{3}=2$ and $2 a_{3}+b_{3}=3$
$\Rightarrow \quad 2(2)+b_{3}=3$
$\Rightarrow \quad 4+b_{3}=3$
$\Rightarrow \quad b_{3}=-1$
and $3 a_{3}+2 b_{3}+c_{3}=1$
$\Rightarrow \quad 3 \times 2+2(-1)+c_{3}=1$
$\Rightarrow \quad 6-2+c_{3}=1$
$\Rightarrow$
$c_{3}=-3$
$\therefore \quad U=\left(\begin{array}{ccc}1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3\end{array}\right)$
$\Rightarrow|U|=1(3-4)-2(6+1)+2(8+1)$
$=-1-14+18=3$
Also, adj $(U)=\left[\begin{array}{ccc}-1 & -7 & 9 \\ -2 & -5 & 6 \\ 0 & -3 & 3\end{array}\right]^{T}=\left[\begin{array}{ccc}-1 & -2 & 0 \\ -7 & -5 & -1 \\ 9 & 6 & 3\end{array}\right]$
$\therefore$
$U^{-1}=\frac{1}{|U|}$ adj $(U)$
$\Rightarrow$
$$
U^{-1}=\left[\begin{array}{rrr}
-\frac{1}{3} & -\frac{2}{3} & 0 \\
-\frac{7}{3} & -\frac{5}{3} & -1 \\
3 & 2 & 1
\end{array}\right]
$$
Hence, the sum of all elements of $U^{-1}$ is $0 .$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2015