Solve this
Question:

Let $X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right], A=\left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A X=B$, then $X$ is equal to

(a)$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

(b) $\left[\begin{array}{r}-1 \\ -2 \\ -3\end{array}\right]$

(c) $\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$

(d) $\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$

(e) $\left[\begin{array}{l}0 \\ 2 \\ 1\end{array}\right]$

Solution:

(a) $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

Here,

$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right], X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$    (Given)

$|A|=1(0-2)+1(2-3)+2(4-0)$

$=-2-1+8$

$=5$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}0 & 1 \\ 2 & 1\end{array}\right|=-2, C_{12}=(-1)^{1+2}\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|=1, C_{13}=(-1)^{1+3}\left|\begin{array}{ll}2 & 0 \\ 3 & 2\end{array}\right|=4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-1 & 2 \\ 2 & 1\end{array}\right|=5, C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & 2 \\ 3 & 1\end{array}\right|=-5, C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right|=-5$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right|=-1, C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right|=3, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & -1 \\ 2 & 0\end{array}\right|=2$

$\operatorname{adj} A=\left[\begin{array}{ccc}-2 & 1 & 4 \\ 5 & -5 & -5 \\ -1 & 3 & 2\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{ccc}-2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{5}\left[\begin{array}{ccc}-2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2\end{array}\right]$

$\therefore X=A^{-1} B$

$\Rightarrow X=\frac{1}{5}\left[\begin{array}{ccc}-2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2\end{array}\right]\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$

$\Rightarrow X=\frac{1}{5}\left[\begin{array}{c}-6+5-4 \\ 3-5+12 \\ 12-5+8\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}-5 \\ 10 \\ 15\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right]$

Administrator

Leave a comment

Please enter comment.
Please enter your name.