How to solve the determinant without expanding?

Question

Here is the the determinant equation: $$ \begin{vmatrix} 1 & bc & bc(b+c)\\ 1 & ca & ca(c+a)\\ 1 & ab & ab(a+b)\\ \end{vmatrix} = abc\begin{vmatrix} a & 1 & b+c\\ b & 1 & c+a\\ c & 1 & a+b\\ \end{vmatrix}=0 $$ I am new and beginner in matrices and determinants.So please explain the steps of the answers.

Answer 1

It's a famous rule of determinants that

If we multiply a row (column) of a matrix A by a number, the determinant of A will be multiplied by the same number.

Suppose we are comfortably in $\mathbb{R}$ and suppose also that $a,b,c \ne 0$, then multiply and divide the first row for $a$, the second for $b$ and the third for $c$ $$ \frac{abc}{abc}\begin{vmatrix} 1 & bc & bc(b+c)\\ 1 & ca & ca(c+a)\\ 1 & ab & ab(a+b)\\ \end{vmatrix} = \frac{1}{abc}\begin{vmatrix} a & abc & abc(b+c)\\ b & abc & abc(c+a)\\ c & abc & abc(a+b)\\ \end{vmatrix} $$

Now you can conclude beacuse taking out the $(abc)^2$ factor that multiplies the second and third column you get

$$ abc\begin{vmatrix} a & 1& (b+c)\\ b & 1 & (c+a)\\ c & 1 & (a+b)\\ \end{vmatrix} $$

EDIT

At this point, to conclude that the determinant is $0$ you can simply use the Sarrus' rule or, without expanding, you can subtract the first row to the second and third row to get

$$ abc \begin{vmatrix} a & 1 & b+c\\ b-a & 0 & a-b\\ c-a & 0 & a-c\\ \end{vmatrix} = abc[(a-b)(c-a) - (a-c)(b-a)] = 0 $$

Notice that in the header of the proof I supposed that $a,b,c \ne 0$, but now we can conclude even if one between $a,b,c = 0$, because the property is satisfied as well.

Answer 2

We have $$\begin{vmatrix} a & 1 & b+c\\ b & 1 & c+a\\ c & 1 & a+b\\ \end{vmatrix} = \begin{vmatrix} a & 1 & a+b+c\\ b & 1 & b+c+a\\ c & 1 & c+a+b\\ \end{vmatrix} = 0$$ Here the first columm was added to the third. Thus it remains to verify the first equation in your question.

Case 1: $abc = 0$.

In that case at least one of $a, b ,c$ is $0$ and $$ \begin{vmatrix} 1 & bc & bc(b+c)\\ 1 & ca & ca(c+a)\\ 1 & ab & ab(a+b)\\ \end{vmatrix}$$ has at least two rows of the form $(1 \quad 0 \quad 0)$ so that the first equation follows.

Case 2: $abc \ne 0$.

Then $$ abc \begin{vmatrix} 1 & bc & bc(b+c)\\ 1 & ca & ca(c+a)\\ 1 & ab & ab(a+b)\\ \end{vmatrix} = \begin{vmatrix} a & abc & abc(b+c)\\ b & abc & abc(c+a)\\ c & abc & abc(a+b)\\ \end{vmatrix} = (abc)^2\begin{vmatrix} a & 1 & b+c\\ b & 1 & c+a\\ c & 1 & a+b\\ \end{vmatrix} $$ In the first step the factor $abc$ was shifted into the determinant by multiplying the first row with $a$, the second with $b$ and the third with $c$. In the second step the factor (abc) was extracted from columns two and three.

Dividing by $abc \ne 0$ yields the desired result.

Answer 3

By gaining knowledge from the above answers I think I have figure it out.Please check my answer if I am wrong correct my answer please.

$$ \begin{vmatrix} 1 & bc & bc(b+c)\\ 1 & ca & ca(c+a)\\ 1 & ab & ab(a+b)\\ \end{vmatrix} = abc\begin{vmatrix} a & 1 & b+c\\ b & 1 & c+a\\ c & 1 & a+b\\ \end{vmatrix} $$ On the left hand side $Row1*a$, $Row2 * b$, $Row3*c$, On the right hand side $Col'3 =Col1 + Col3$ $$ \begin{vmatrix} a & abc & abc(b+c)\\ b & abc & abc(c+a)\\ c & abc & abc(a+b)\\ \end{vmatrix} =abc\begin{vmatrix} a & 1 & a+b+c\\ b & 1 & a+b+c\\ c & 1 & a+b+c\\ \end{vmatrix} $$ Now on the left Hand Side factoring $abc$ and On the right hand side factor $a+b+c$. $$abc^2\begin{vmatrix} a & 1 & b+c\\ b & 1 & c+a\\ c & 1 & a+b\\ \end{vmatrix}=abc(a+b+c)\begin{vmatrix} a & 1 & 1\\ b & 1 & 1\\ c & 1 & 1\\ \end{vmatrix} $$ Now according to the determinant rule, if a determinant has two identical columns then it's value is 0.Which means the right side is 0. And for the left hand side $Col'3 = Col1 + Col3$ and factor a+b+c which eventually lead to the same situation. $$abc^2\begin{vmatrix} a & 1 & a + b + c\\ b & 1 & a+b+c\\ c & 1 & a+b+c\\ \end{vmatrix}= abc(a+b+c)*0 $$ $$ abc^2(a+b+c)\begin{vmatrix} a & 1 &1\\ b & 1 &1\\ c & 1 & 1\\ \end{vmatrix}= 0 $$ $abc^2(a+b+c)*0 = 0$

$0=0=0$