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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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