# Applications Of Determinants To Linear Equations

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Consider a system of linear equations in two variables:

\begin{align} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0 \\\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0 \\ \end{align}

Graphically, these represent two lines which may be parallel, or which may intersect at a unique point.

If we solve this system, we obtain

\begin{align} & \qquad \quad \frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{-y}{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\ \\ & \Rightarrow \qquad \frac{x}{\left| \ \begin{matrix} {{b}_{1}} & {{c}_{1}} \\ {{b}_{2}} & {{c}_{2}} \\\end{matrix}\ \right|}=\frac{-y}{\left| \ \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\\end{matrix}\ \right|}=\frac{1}{\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|} \\ \\ & \Rightarrow \qquad x=\frac{\left| \ \begin{matrix} {{b}_{1}} & {{c}_{1}} \\ {{b}_{2}} & {{c}_{2}} \\\end{matrix}\ \right|}{\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|}\ \ ,\ \ y=\frac{-\left| \ \begin{matrix} {{a}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{c}_{2}} \\\end{matrix}\ \right|}{\ \ \left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|} \qquad \qquad \ldots (1) \\ \end{align}

If $$\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|\ne 0$$ , then a unique solution will exist for x and y physically,  $$\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|\ne 0$$ means that \begin{align}{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}\ne 0,\ \text{or}\ \frac{{{a}_{1}}}{{{b}_{1}}}\ne \frac{{{a}_{2}}}{{{b}_{2}}} \end{align} , which implies that the two lines are non-parallel, so a unique solution must exist.

If $$\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} \\{{a}_{2}} & {{b}_{2}} \\\end{matrix}\ \right|=0$$, the two lines are parallel, because  \begin{align}\frac{{{a}_{1}}}{{{b}_{1}}}=\frac{{{a}_{2}}}{{{b}_{2}}}. \end{align} In this case, if \begin{align}\frac{{{c}_{1}}}{{{c}_{2}}}\ne \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\end{align}, the two lines are different and parallel, so no solution exists; this is confirmed in (1) because the denominator-determinant is 0, but the numerator-determinants are not.

The other case is when \begin{align}\frac{{{c}_{1}}}{{{c}_{2}}}\ne \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}; \end{align} this means that in (1), all the determinants will be 0. Physically, this means that the two lines are the same, so there will be infinitely many solutions.

Let us now consider a system in three variables:

\begin{align} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0 \\\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0 \\\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z+{{d}_{3}}=0 \\ \end{align}

Physically, this corresponds to three planes in 3-D space. If the three planes intersect in a point, a unique solution will exist for the system. But the three planes may not intersect in a point. For example, if two of the planes are parallel, there is no unique intersection point. In such cases, the system does not have a solution. Also, there will be cases where infinitely many solutins exist. For example, when the third plane passes throught the line of intersection of the Arst two planes, the set of intersection points is the same line, which means there are infinitely many solutions.

Let us now determine the solutions ot this system. Taking & cue from from the result of the two-variables system, we define

$\Delta =\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix}\ \right|\ \ \text{and}\ {{\Delta }_{1}}=\left| \ \begin{matrix} {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\\end{matrix}\ \right|$

Now, on $${{\Delta }_{1}},$$ if we apply the transformation  $${{C}_{3}}\to y{{C}_{1}}+z{{C}_{2}}+{{C}_{3}},$$ we obtain

${{\Delta }_{1}}=\left| \ \begin{matrix} {{b}_{1}} & {{c}_{1}} & {{b}_{1}}y+{{c}_{1}}z+{{d}_{1}} \\ {{b}_{2}} & {{c}_{2}} & {{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \\ {{b}_{3}} & {{c}_{3}} & {{b}_{3}}y+{{c}_{3}}z+{{d}_{3}} \\\end{matrix}\ \right|$

But using the original equatons, the third column C3 reduces to

\begin{align} {{\Delta }_{1}}&=\left| \ \begin{matrix} {{b}_{1}} & {{c}_{1}} & -{{a}_{1}}x \\ {{b}_{2}} & {{c}_{2}} & -{{a}_{2}}x \\ {{b}_{3}} & {{c}_{3}} & -{{a}_{3}}x \\\end{matrix}\ \right| \\ \\ & =-x\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix}\ \right| \\\\ & =-x\Delta \\\\ \Rightarrow \quad x&=-\frac{{{\Delta }_{1}}}{\Delta } \\ \end{align}

Using an exactly analogour process, we can arrive at expressions for y and z:

\begin{align} & y=\frac{+{{\Delta }_{2}}}{\Delta }\ \ \ \text{where}\ \ \ {{\Delta }_{2}}=\left| \ \begin{matrix} {{a}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{a}_{2}} & {{c}_{3}} & {{d}_{3}} \\\end{matrix}\ \right| \\ \\ & z=\frac{-{{\Delta }_{3}}}{\Delta }\ \ \ \text{where}\ \ \ {{\Delta }_{3}}=\left| \ \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{c}_{3}} & {{d}_{3}} \\\end{matrix}\ \right| \\ \end{align}

Therefore, we arrive at what is known as the Cramer’s rule:

$\boxed {\frac{x}{{{\Delta }_{1}}}=\frac{-y}{{{\Delta }_{2}}}=\frac{z}{{{\Delta }_{3}}}=\frac{-1}{\Delta }}$

As is the two variables system, if $$\Delta \ne 0,$$ then obviously a unique solution will exist for $$\left( x,y,z \right)$$, given by the Cramer’s rule. That solution will physically correspond to the point of intersection of the three planes represented by the three equations of the system.

However, if $$\Delta =0,$$ then two cases may arise: in case all the other determinants  $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ are also zero, then the system will have infinite solutions. However, if even one of $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ is non-zero, then the system will have no solution, because Cramer’s rule will give rise to undefined quantities.

Let us summarize:

 $$\Delta \ne 0,$$ $$\Delta =0$$ $$\to$$    Unique solution exists $$\to$$    If $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ are all zero, infinite solutions exist $$\to$$    If at least one of  $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ is non-zero, no solution exists
Determinants and Matrices
Determinants and Matrices
grade 11 | Questions Set 1
Determinants and Matrices
Determinants and Matrices
grade 11 | Questions Set 2