Question

# Without expanding at any stage show that$$\begin{vmatrix} { x }^{ 2 }+x & x+1 & x-2 \\ 2{ x }^{ 2 }+3x-1 & \qquad 3x & 3x-3 \\ { x }^{ 2 }+2x+3 & 2x-1 & 2x-1 \end{vmatrix}=xA+B$$where $$A$$ and $$B$$ are determinants of order $$3$$ not involving $$x$$.

Solution

## Apply $$R_{ 2 }-(R_{ 1 }+R_{ 3 })$$$$\Delta =\left| \begin{matrix} { x }^{ 2 }+x & x+1 & x-2 \\ -4 & 0 & 0 \\ { x }^{ 2 }+2x+3 & 2x-1 & 2x-1 \end{matrix} \right|$$Now apply $$R_{ 1 }+\dfrac { 1 }{ 4 } { x }^{ 2 }{ R }_{ 2 },{ R }_{ 3 }+\dfrac { 1 }{ 4 } { x }^{ 2 }{ R }_{ 2 }$$$$\Delta =\left| \begin{matrix} x & x+1 & x-2 \\ -4 & 0 & 0 \\ 2x+3 & 2x-1 & 2x-1 \end{matrix} \right|$$Apply $$R_{ 3 }-2R_{ 1 }$$$$\Delta =\left| \begin{matrix} x+0 & x+1 & x-2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{matrix} \right|$$$$\Delta =\left| \begin{matrix} x & x & x \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{matrix} \right| +\left| \begin{matrix} 0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{matrix} \right|$$$$=x\left| \begin{matrix} 1 & 1 & 1 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{matrix} \right| +\left| \begin{matrix} 0 & 1 & -2 \\ -4 & 0 & 0 \\ 3 & -3 & 3 \end{matrix} \right|$$which is of the form $$x\ A+B$$.Mathematics

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