Question

For a matrix
$A=\begin{pmatrix}1&0&0\\ 2&1&0\\ 3&2&1\end{pmatrix},$ if $U_{1}, U_{2}$ and $U_{3}$ are $3\times1$ column matrices satisfying
$AU_{1}=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix}, AU_{2}=\begin{pmatrix}2\\ 3\\ 0\end{pmatrix}, AU_{3}=\begin{pmatrix}2\\ 3\\ 1\end{pmatrix}$ and $U$ is $3\times3$ matrix whose columns are $U_1, U_2$ and $U_3$
Then sum of the elements of $U^{-1}$ is

• A. 6
• B. 0
• C. 1
• D. 2/3

Answer 1

Answer: (B)

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