value of $a,b,c,d,e$ in expansion of determinant

Question

If $\displaystyle \begin{vmatrix}x^2+3x & x-1 & x+3 \\ x+1 & -2x & x-4 \\ x-3 & x+4 & 3x\end{vmatrix}=ax^4+bx^3+cx^2+dx+e$

Then $a,b,c,d,e$ equals

I have tried several times using transformation properties of determinant but I could not get any satisfactory way

So here I have expand along $(1)$ st Row

$\displaystyle (x^2+3x)\bigg[(-2x)(3x)-(x^2-16)\bigg]-(x-1)\bigg[(3x)(x+1)-(x-4)(x-3)\bigg]+(x+3)\bigg[(x+1)(x+4)+(2x)(x-3)\bigg]$

$\displaystyle (x^2+3x)(16-7x^2)-(x-1)(2x^2+10x-12)+(x+3)(3x^2-x+4)$

$\displaystyle -7x^4-20x^3+36x^2+27x+24$

After comparing, we get

$a=-7,b=-20,c=36,d=27,e=24$

How can We expand the above determinant using transformation properties of determinant , please explain me, Thanks

Answer 1

I think there is an error in the step from

$(x^2+3x)(16-7x^2)-(x-1)(2x^2+10x-12)+(x+3)(3x^2-x+4)$

to

$-7x^4-20x^3+36x^2+27x+24$

---

Although this answer does not use transformation properties of determinant, this answer shows some observations which should make the calculations easier.

Some observations :

• $a_{11}=x(x+3),a_{22}=-2x,a_{33}=3x$ are divisible by $x$.

• $a_{12}a_{23}a_{31}=(x\color{red}{-1})(x\color{red}{-4})(x\color{red}{-3})$

• $a_{21}a_{32}a_{13}=(x\color{red}{+1})(x\color{red}{+4})(x\color{red}{+3})$

• Since $(x\pm a)(x\pm b)(x\pm c)=x^3\pm(a+b+c)x^2+(ab+bc+ca)x\pm abc$, we see that $$a_{12}a_{23}a_{31}+a_{21}a_{32}a_{13}=2x^3+2(1\cdot 4+4\cdot 3+3\cdot 1)x=x(2x^2+38)$$ is divisible by $x$.

By the Rule of Sarrus, the determinant is given by $$\begin{align}&\underbrace{a_{12}a_{23}a_{31}+a_{21}a_{32}a_{13}}_{\text{divisible by $x$}}+\underbrace{\color{red}{a_{11}a_{22}a_{33}}-\color{red}{a_{22}}a_{13}a_{31}-\color{red}{a_{33}}a_{12}a_{21}-\color{red}{a_{11}}a_{23}a_{32}}_{\text{divisible by $x$}} \\\\&=x\bigg(2x^2+38+(x+3)\times (-2x)\times 3x-(-2)(x+3)(x-3)-3(x+1)(x-1) \\&\qquad\qquad-(x+3)(x+4)(x-4)\bigg) \\\\&=x\bigg(2x^2+38-6x^2(x+3)+2(x^2-9)-3(x^2-1)-(x+3)(x^2-16)\bigg) \\\\&=x\bigg((-6-1)x^3+(2-18+2-3-3)x^2+16x+38-18+3+48\bigg) \\\\&=x(-7x^3-20x^2+16x+71) \\\\&=-7x^4-20x^3+16x^2+71x\end{align}$$

Therefore, the answer is $$\color{red}{a=-7,\quad b=-20,\quad c=16,\quad d=71,\quad e=0}$$

Answer 2

We can decompose the determinant by the first column: $$ \begin{vmatrix} x^2+3x & x-1 & x+3\\ x+1 & -2x & x-4\\ x-3 & x+4 & 3x \end{vmatrix}$$ $$=(x^2+3x)\begin{vmatrix} -2x & x-4\\ x+4 & 3x \end{vmatrix} -(x+1)\begin{vmatrix} x-1 & x+3\\ x+4 & 3x \end{vmatrix} +(x-3)\begin{vmatrix} x-1 & x+3\\ -2x & x-4\\ \end{vmatrix} $$ $$=(x^2+3x)(-7x^2+16)-(x+1)(2x^2-10x-12)+(x-3)(3x^2+x+4)$$ $$=-7x^4+(-21-2+3)x^3+(16-2+10-9+1)x^2+(48+10+12+4-3)x+12-12$$ $$=-7x^4-20x^3+16x^2+71x.$$ Also, we can use the Sarrus' rule: $$ \begin{vmatrix} x^2+3x\color{brown}{_{\LARGE_\searrow}}\hspace{-20mu} & x-1\color{brown}{_{\LARGE_\searrow}} & x+3\color{brown}{_{\LARGE_\searrow}} & x^2+3x & x-1\\ x+1 & -2x\color{green}{^{\LARGE_\nearrow}}\hspace{-16mu}\color{brown}{_{\LARGE_\searrow}}\hspace{-16mu} & x-4\color{green}{^{\LARGE_\nearrow}}\hspace{-16mu}\color{brown}{_{\LARGE_\searrow}} &x+1\color{green}{^{\LARGE_\nearrow}}\hspace{-16mu}\color{brown}{_{\LARGE_\searrow}}\hspace{-20mu} & -2x\\ x-3\color{green}{^{\LARGE_\nearrow}} & x+4\color{green}{^{\LARGE_\nearrow}}\hspace{-20mu} & 3x\color{green}{^{\LARGE_\nearrow}}\hspace{-20mu} & x-3 & x+4 \end{vmatrix}$$ $$(x^2+3x)(-2x)\cdot 3x +(x-1)(x-4)(x-3)+(x+3)(x+1)(x+4)$$ $$-(x-3)(-2x)(x+3)-(x+4)(x-4)(x^2+3x)-3x\cdot(x+1)(x-1)$$ $$=-6x^2(x^2+3x)+2(x^3+(1\cdot4+4\cdot3+3\cdot1)x)$$ $$+2x(x^2-9)-(x^2+3x)(x^2-16)-3x(x^2-1)$$ $$=-6x^3(x+3)+2(x^3+19x)$$ $$+2x(x^2-9)-(x^2+3x)(x^2-16)+3x(1-x^2)$$ $$=(-6-1)x^4+(-18+2+2-3-3)x^3+16x^2+(38-18+48+3)x$$ $$=-7x^4-20x^3+16x^2-71x.$$

Answer 3

Here is yet an other way to go, the first step is using the linearity of the determinant w.r.t. the first line, which is $(x^2+3x,x-1,x+3)=(x^2+3x, x,x)+(0,-1,3)$. The determinant $\Delta$ to be computed is thus $\Delta =\Delta_1+\Delta_2$ where $$ \begin{aligned} \Delta_1 &= \begin{vmatrix} x^2+3x & x & x\\ x+1 & -2x & x-4\\ x-3 & x+4 & 3x \end{vmatrix} = x \begin{vmatrix} x+3 & 1 & 1\\ x+1 & -2x & x-4\\ x-3 & x+4 & 3x \end{vmatrix} \\ &=x(x+3) \begin{vmatrix} -2x & x-4\\ x+4 & 3x \end{vmatrix} - x \begin{vmatrix} x+1 & x-4\\ x-3 & 3x \end{vmatrix} +x \begin{vmatrix} x+1 & -2x \\ x-3 & x+4 \end{vmatrix} \\ &=x(x+3)(-7x^2+16)-x(2x^2+10x-12)+x(3x^2-x+4) \\ &=-x(7x^3 + 20x^2 - 5x - 64)\ , \\[3mm] \Delta_2 &= \begin{vmatrix} 0 & \boxed{-1} & +3\\ x+1 & -2x & x-4\\ x-3 & x+4 & 3x \end{vmatrix}= \begin{vmatrix} 0 & -1 & 0\\ x+1 & * & -5x-4\\ x-3 & * & 6x+12 \end{vmatrix}= \begin{vmatrix} x+1 & -5x-4\\ x-3 & 6x+12 \end{vmatrix} \\ &= \begin{vmatrix} x+1 & -5x-4\\ -4 & 11x+16 \end{vmatrix}= \begin{vmatrix} x+1 & 1\\ -4 & 11x-4 \end{vmatrix} =11x^2+7x=x(11x+7)\ ,\text{ so} \\[3mm] \Delta&=\Delta_1+\Delta_2=-x\Big(\ (7x^3 + 20x^2 - 5x - 64)-(11x+7)\ \Big) \\ &=-x(7x^3 + 20x^2 - 16x - 71)\ . \end{aligned} $$ The needed coefficients $a,b,c,d,e$ can now be easily extracted, they are respectively $$\bbox[yellow]{-7\ , \ -20\ ,\ 16\ ,\ 71\ , \ 0\ .}$$

Answer 4

Given

$$ \text{det} = \begin{vmatrix}x^2+3x & x-1 & x+3 \\ x+1 & -2x & x-4 \\ x-3 & x+4 & 3x\end{vmatrix}=ax^4+bx^3+cx^2+dx+e, $$

expand along the first row to get

$$ \text{det} = (x^2+3x)\bigg[(-2x)(3x)-(x^2-16)\bigg] \\ -(x-1)\bigg[(3x)(x+1)-(x-4)(x-3)\bigg] \\ +(x+3)\bigg[(x+1)(x+4)+(2x)(x-3)\bigg].$$

Expand the bracketed terms to get

$$ \text{det} = (x^2+3x)(16-7x^2) -(x-1)(2x^2+10x-12)+(x+3)(3x^2-x+4) .$$

Multiply out each term to get

$$ \text{det} = (-7 x^4-21 x^3+16 x^2+48 x) - (2 x^3+8 x^2-22 x+12 ) + (3 x^3+8 x^2+x+12 ). $$

Combine the polynomial terms together to get

$$ \text{det} = -7 x^4-20 x^3+16 x^2+71 x. $$

Hence,

$a=-7,b=-20,c=16,d=71,e=0.$

There are other ways to calculate a $3\times 3$ determinant but this seems to be the easiest to work with.