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Considering three equations $$\begin{cases}a_1x+b_1y+c_1z=d_1\\a_2x+b_2y+c_2z=d_2\\a_3x+b_3y+c_3z=d_3\end{cases}$$ let $$\Delta=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} \Delta_x=\begin{vmatrix}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{vmatrix} \Delta_y=\begin{vmatrix}a_1&d_1&c_1\\a_2&d_2&c_2\\a_3&d_3&c_3\end{vmatrix} \Delta_z=\begin{vmatrix}a_1&b_1&d_1\\a_2&b_2&d_2\\a_3&b_3&d_3\end{vmatrix}$$ if the three planes intersect in a line, then $\Delta=\Delta_x=\Delta_y=\Delta_z=0$

I know the reason why $\Delta=0$, but I don't understand why$\Delta_x=\Delta_y=\Delta_z=0$

My thought is that if one of them doesn't equal to $0$, then the equations will have no solution, so they must equal to $0$

But can we explain that by using properties of determinant?

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2 Answers 2

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Multiply the given three eqns. by $\vec i, \vec j, \vec k$ we get $$\vec A x+ \vec B y +\vec C z= \vec D,~~ \vec A=a_1\vec i+ a_2 \vec j+ a_3 \vec k, etc. $$ Take cross multiplication of this eqn. by $(\vec B \times \vec C)$ from left to get $$[\vec A, \vec B, \vec C]x=[\vec D, \vec B,\vec C]\implies \Delta x= \Delta_1$$ Hwre $[\vec P, \vec Q, \vec R]$ is the scalar (box) product of three vectors which is alwaysa determinant. Similarly, we get $\Delta y= \Delta_2$ and $\Delta z=\Delta_3$ so $\Delta=\Delta_1=\Delta_2=\Delta_3=0$ lead to a consistent result $0=0$ three times Hence the three equation are consistent and they will have at least one solution. There will be only one (unique) solution if $\Delta \ne 0$, other wise many solutions. If $\Delta=0$ but any one of $\Delta_1, \Delta_2, \Delta_3$ is non-xero, it will lead to inconsistency so no solution.

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Consider the matrix $$ A = \begin{bmatrix} \vec{a} & \vec{b} & \vec{c} \\ \end{bmatrix} $$

$$ K = \begin{bmatrix} \textbf{A} & \vec{d} \\ \end{bmatrix} $$ This Matrix has $\text{dim}$ $\mathbb{N}(A) = 1$, therefore it's column space has $\text{dim}\ \text{Col}(A) = 2$. I.e two rows are linearly independent. If $\Delta_x$ was not $0$ then it would follow that $\vec{d}$ is independent of $\vec{b}$ and $\vec{c}$ and that $b$ and $c$ are the Linearly independent of each other. This also implies that $a$ must be written as a combination of $b$ and $c$. since $dim\ \text{col} (A) = 2$. But since $\textbf{d}$ belongs to the column space of $A = [ \vec{a} \ \ \vec{b} \ \ \vec{c} ]$ it follows that the $\vec{d}$ is also expressible as a combination of $\vec{b}$ and $\vec{c}$ , which is a contradiction to $\Delta_x = 0$

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