While solving a system of linear equations $\mathrm{AX}=\mathrm{B}$ using Cramer's rule with the usual…

While solving a system of linear equations $\mathrm{AX}=\mathrm{B}$ using Cramer's rule with the usual notation if $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right| ; \Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5\end{array}\right|$ and $X=\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]$ then $\alpha^2+\beta^2=$
  1. $9$
  2. $13$
  3. $5$
  4. $25$

Solution

Given, $\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right| \Rightarrow \Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5\end{array}\right|$ and $x=\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]$ Now, $A X=B \Rightarrow\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right]\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]=\left[\begin{array}{l}5 \\ 4 \\ 11\end{array}\right]$ So, $\alpha+2+\beta=5 \Rightarrow \alpha+\beta=3 \ldots(i)$ and, $-\alpha+2+5 \beta=11 \Rightarrow-\alpha+5 \beta=9 \ldots(ii)$ After solving equations (i) and (ii), we get $\alpha=1, \beta=2$ So, $\alpha^2+\beta^2=1+4=5$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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