Question

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
$A{U}_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),A{U}_2=\left(\begin{array}{c}2\\ 3\\ 0\end{array}\right),A{U}_3=\left(\begin{array}{c}2\\ 3\\ 1\end{array}\right)$ and $U$ is $3\times 3$ matrix whose columns are $U_1,U_2$ and $U_3$
Then sum of the elements of $U^{-1}$ is

• A. 6
• B. 0
• C. 1
• D. 2/3

Answer 1

Answer: (B)

Let $U_1=\left(\begin{array}{c}{a}_i\\ b_i\\ c_i\end{array}\right)$, where $i=1,2,3$
$\therefore A{U}_1=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right)\left(\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right)$
$\Rightarrow \left(\begin{array}{c}{a}_1\\ 2a_1+b_1\\ 3a_1+2b_1+c_1\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$
$\left[\because A{U}_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\right]$
$\therefore a_1=1,2a_1+b_1=0$
$\Rightarrow 2+b_1=0$
$\Rightarrow b_1=-2$
and $3a_1+2b_1+c_1=0$
$\Rightarrow 3+2(-2)+c_1=0$
$\Rightarrow 3-4+c_1=0$
$\Rightarrow c_1=1$
Similarly, $A{U}_2=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right)\left(\begin{array}{c}{a}_2\\ b_2\\ c_2\end{array}\right)$
$\Rightarrow \left(\begin{array}{c}{a}_2\\ 2a_2+b_2\\ 3a_2+2b_2+c_2\end{array}\right)=\left(\begin{array}{c}2\\ 3\\ 0\end{array}\right)$
$\left[\because A{U}_2=\left(\begin{array}{c}2\\ 3\\ 0\end{array}\right)\right]$
$\therefore a_2=2$ and $2a_2+b_2=3$
$\Rightarrow 2\times 2+b_2=3$
$\Rightarrow 4+b_2=3$
$\Rightarrow b_2=-1$
and $3a_2+2b_2+c_2=0$
$\Rightarrow 3\times 2+2(-1)+c_2=0$
$\Rightarrow 6-2+c_2=0$
$\Rightarrow c_2=-4$
Now, $A{U}_3=\left(\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right)\left(\begin{array}{c}{a}_3\\ b_3\\ c_3\end{array}\right)$
$\Rightarrow \left(\begin{array}{c}{a}_3\\ 2a_3+b_3\\ 3a_3+2b_3+c_3\end{array}\right)=\left(\begin{array}{c}2\\ 3\\ 1\end{array}\right)$
$\left[\because A{U}_3=\left(\begin{array}{c}2\\ 3\\ 1\end{array}\right)\right]$
$\therefore a_3=2$ and $2a_3+b_3=3$
$\Rightarrow 2(2)+b_3=3$
$\Rightarrow 4+b_3=3$
$\Rightarrow b_3=-1$
and $3a_3+2b_3+c_3=1$
$\Rightarrow 3\times 2+2(-1)+c_3=1$
$\Rightarrow 6-2+c_3=1$
$\Rightarrow c_3=-3$
$\Rightarrow U=\left(\begin{array}{cccc}1& 2& 2& \\ -2& -1& -1& \\ 1& -4& -3& \end{array}\right)$
$\Rightarrow |U|=1(3-4)-2(6+1)+2(8+1)$
$=-1-14+18=3$
Also, adj $(U)={\left[\begin{array}{ccc}-1& -7& 9\\ -2& -5& 6\\ 0& -3& 3\end{array}\right]}^T=\left[\begin{array}{ccc}-1& -2& 0\\ -7& -5& -3\\ 9& 6& 3\end{array}\right]$
$\therefore U^{-1}=\frac{1}{|U|}$ adj (U)
$\Rightarrow U^{-1}=\left[\begin{array}{ccc}-\frac{1}{3}& -\frac{2}{3}& 0\\ -\frac{7}{3}& -\frac{5}{3}& -1\\ 3& 2& 1\end{array}\right]$
Hence, the sum of all elements of $U^{-1}$ is $0.$