Matrices And Determinants Ques 66
Without expanding a determinant at any stage, show that
$ \left|\begin{array}{ccc} x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3=x A+B \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1 \end{array}\right| $ $= xA+B$
where $A$ and $B$ are determinants of order 3 not involving $x$.
$(1982,5 \mathrm{M})$
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Solution:
Formula:
- Let $\Delta=\left|\begin{array}{ccr}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-\left(R_{1}+R_{3}\right)$, we get
$ \Delta=\left|\begin{array}{ccc} x^{2}+x & x+1 & x-2 \\ -4 & 0 & 0 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1 \end{array}\right| $
Applying $R_{1} \rightarrow R_{1}+\frac{x^{2}}{4} R_{2}$
and $R_{3} \rightarrow R_{3}+\frac{x^{2}}{4} R_{2}$, we get
$ \Delta=\left|\begin{array}{ccc} x & x+1 & x-2 \\ -4 & 0 & 0 \\ 2 x+3 & 2 x-1 & 2 x-1 \end{array}\right| $
Applying $R_{3} \rightarrow R_{3}-2 R_{1}=\left|\begin{array}{ccc}x+0 & x+1 & x-2 \\ -4 & 0 & 0 \\ 3 & -3 & 3\end{array}\right|$
$ \left|\begin{array}{rrr} x & x & x \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{array} \right| $ + $ \left|\begin{array}{rrr} 0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{array} \right| $
= $x$ $ \left|\begin{array}{rrr} 1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{array} \right| $ + $ \left|\begin{array}{rrr} 0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{array} \right| $
$\Rightarrow \quad \Delta=A x+B$
where, $A=\left|\begin{array}{rrr}1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3\end{array}\right|$
and $B=\left|\begin{array}{rrr}0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3\end{array}\right|$
BETA
