Examples On Applications Of Determinants To Linear Equations

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Example - 16

Solve the system of equations

\[\begin{align}  & 2x+y+z=7 \\\\  & x+2y+3z=14 \\\\  & x-3y+2z=1 \\\\ \end{align}\]

Solution:

\[\begin{align}  & \Delta =\left| \ \begin{matrix}   2 & 1 & 1  \\   1 & 2 & 3  \\   1 & -3 & 2  \\\end{matrix}\  \right|=22\ \ \ {{\Delta }_{1}}=\left| \ \begin{matrix}   1 & 1 & -7  \\   2 & 3 & -14  \\   -3 & 2 & -1  \\\end{matrix}\  \right|=-22 \\ \\  & {{\Delta }_{2}}=\left| \ \begin{matrix}   2 & 1 & -7  \\   1 & 3 & -14  \\   1 & 2 & -1  \\\end{matrix}\  \right|=44\ \ \ {{\Delta }_{3}}=\left| \ \begin{matrix}   2 & 1 & -7  \\   1 & 2 & -14  \\   1 & -3 & -1  \\\end{matrix}\  \right|=-66 \\ \end{align}\]

So,

\[x=\frac{-{{\Delta }_{1}}}{\Delta }=1\ ;\ \ y=\frac{+{{\Delta }_{2}}}{\Delta }=2,\ \ z=\frac{-{{\Delta }_{3}}}{\Delta }=3\]

The solution is (1, 2, 3).

A particular case of the 3-variables system is when all the constant terms are zero. Such a system is called hamogenous system, since all ther terms in each equation are of the same degree, i.e. linear:

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

Note that \(x=y=z=0\) is always a solution of this system. But the system may also have infinite number of solutions, in which case there will exist solutions other than that (0, 0, 0). The (0, 0, 0) solution is called the trivial solution for this system, while other solutions (if they exist) are called non-trivial.

For this system \({{\Delta }_{1}}={{\Delta }_{2}}={{\Delta }_{3}}=0,\), and so 0, we have two possible cases:

\(\Delta \ne 0\) \(\Delta =0\)

The system has the trivial solution (0,0,0) 

and no other solution

The system has non-trivial solutions, because

infinite solutions exist.

Example - 17

Find the values of p and q for which the system

\[\begin{align}  & 2x+py+6z=8 \\\\  & \ \ x+2y+qz=5 \\\\  & \ \ x+\ \ y+3z=4 \\ \end{align}\]

has (a) no solution (b) a unique solution (c) infinite solutions

Solution:

\[\begin{align}  & \Delta =\left| \ \begin{matrix}   2 & p & 6  \\   1 & 2 & q  \\   1 & 1 & 3  \\\end{matrix}\  \right|=\ \left( p-2 \right)\left( q-3 \right) \\ \\
 & {{\Delta }_{1}}=\left| \ \begin{matrix}   p & 6 & -8  \\   2 & q & -5  \\   1 & 3 & -4  \\\end{matrix}\  \right|=\ \left( p-2 \right)\left( 15-4q \right) \\ \\
 & {{\Delta }_{2}}=\left| \ \begin{matrix}   2 & 6 & -8  \\   1 & q & -5  \\   1 & 3 & -4  \\\\\end{matrix}\  \right|=\ 0 \\ \\\text{and} \quad  & {{\Delta }_{3}}=\left| \ \begin{matrix}   2 & p & -8  \\   1 & 2 & -5  \\   1 & 1 & -4  \\\end{matrix}\  \right|=\ 2-p \\ \end{align}\]

(a) For no solution \(\Delta =0\) and at least one of \({{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}\) should be non-zero. This can happen if

\[q=3,\ \ p\ne 2\]

(b) For a unique solution, \(\Delta \ne 0:\)

\[p\ne 2,\ \ q\ne 3\]

(c) For infinitely many solutions, \(\Delta ={{\Delta }_{1}}={{\Delta }_{2}}={{\Delta }_{3}}=0:\)

\[p=2\]

No constraint on q is required.

Example - 18

Consider the homogenous system

\[\begin{align}  & \ \ x-cy\ -bz=0 \\\\  & \ cx-y\ +az=0 \\ \\ & \ bx+ay\ -z=0 \\ \end{align}\]

If this system has a non-trivial solution, show that

\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\]

Solution: If homogenous system has non-trivial solutions, this implies that \(\Delta =0:\)

\[\left| \ \begin{matrix}   1 & -c & -b  \\   c & -1 & a  \\   b & a & -1  \\\end{matrix}\  \right|=0\]

Using \({{C}_{2}}\to {{C}_{2}}+c{{C}_{1}}\ \ \ \text{and}\ \ {{C}_{3}}\to {{C}_{3}}+b{{C}_{1}},\) we have

\[\begin{align}\Delta   =& \left| \ \begin{matrix}   1 & 0 & 0  \\   c & -1+{{c}^{2}} & a+bc  \\   b & a+bc & -1+{{b}^{2}}  \\\end{matrix}\  \right|=0 \\ \\ 
  \Rightarrow \qquad & \left( -1+{{c}^{2}} \right)\left( -1+{{b}^{2}} \right)-{{\left( a+bc \right)}^{2}}=0 \\ \end{align}\]

upon simplifying, we have

\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\]

Example - 19

Prove that the system of equations

\[\begin{align}  & 3x-y+4z=3 \\\\  & x+2y-3z=-2 \\ \\ & 6x+5y+\lambda z=-3 \\ \end{align}\]

has at least one solution for any real \(\lambda \). Find the solution(s) for \(\lambda =-5.\)

Solution:

\[\Delta =\left| \ \begin{matrix}   3 & -1 & 4  \\   1 & 2 & -3  \\   6 & 5 & \lambda   \\\end{matrix}\  \right|=7\left( \lambda +5 \right)\]

If  \(\lambda \ne -5,\ \ \Delta \ne 0\) so a unique solution exists

\(\mathbf{For}\ \mathbf{\lambda =}\ \mathbf{5}\)

\[{{\Delta }_{1}}=\left| \ \begin{matrix}   -1 & 4 & -3  \\   2 & -3 & 2  \\   5 & -5 & 3  \\\end{matrix}\ \right|=0\]

Similarly,  \({{\Delta }_{2}}={{\Delta }_{3}}=0.\)

That is, for \(\lambda =-5,\ \ \Delta ={{\Delta }_{1}}={{\Delta }_{2}}={{\Delta }_{3}}=0,\) so in this case infinitely many solutions exist. Therefore, in all cases, the system has at least one solution.

Now, how do we express the set of infinite solutions in the case of  \(\lambda =-5?\) What we have to realize is that if we express two of the variables in terms of the third, and vary the third as a parameter, we get all the possible solutions. For \(\lambda =-5,\) abserve that y and z can be written in terms of x as

\[y=\frac{1-13x}{5},z=\frac{4-7x}{5} \qquad \left( \text{Verify}! \right)\]

As x is varied, y and z vary in a manner such that (x, y, z) is a valid solution to the given system. Therefore, the set of all possible solutions can be expressed as

\[\left( x=\frac{1-13x}{5},\frac{4-7x}{5} \right),\ x\in \mathbb{R}\]

TRY YOURSELF - III

Q. 1    If  \(f\left( x \right)=a{{x}^{2}}+bx+c\ \ \text{and}\ f\left( -2 \right)=11,\ f\left( 0 \right)=3,\ \ f\left( 1 \right)=2,\) determine \(f\left( x \right)\). Use Cramer’s rule.

Q. 2    If \(x,y,z\) are not all zero and if

\[\begin{align}  & ax+by+cz=0 \\\\  & bx+cy+az=0 \\\\  & cx+ay+bz=0 \\ \end{align}\]

Find the possible values of the ratio \(x:y:z\) .

Q. 3    Let  \(\lambda \) and \(\alpha \) be real. Find the set of all values of  \(\lambda \) for which the system

\[\begin{align}  & \lambda x+\left( \sin \alpha  \right)y+\left( \cos \alpha  \right)z=0 \\\\  & \ \ x+\left( \cos \alpha  \right)y+\left( \sin \alpha  \right)z=0 \\\\  & -x+\left( \sin \alpha  \right)y-\left( \cos \alpha  \right)z=0 \\ \end{align}\]

has a non-trivial solution. For \(\lambda =1,\) find the possible values of  \(\lambda \).