Question

Solve the given system of linear equation by Crammer’s rule
$x+y+z=1$
$ax+by+cz=k$
${{a}^{2}}x+{{b}^{2}}y+{{c}^{2}}z={{k}^{2}}$.

Answer 1

Hint: We start solving the problem by writing the given system of linear equations in the form of $AX=B$. We then recall the Crammer’s rule that the solution for system of linear equations is $x=\dfrac{\left| {{A}_{1}} \right|}{\left| A \right|}$, $y=\dfrac{\left| {{A}_{2}} \right|}{\left| A \right|}$, $z=\dfrac{\left| {{A}_{3}} \right|}{\left| A \right|}$ and then write the matrices A, ${{A}_{1}}$, ${{A}_{2}}$, ${{A}_{3}}$. We then find the values of the determinants $\left| A \right|$, $\left| {{A}_{1}} \right|$, $\left| {{A}_{2}} \right|$, $\left| {{A}_{3}} \right|$ using the facts that determinant of a $3\times 3$ matrix is $\left| \begin{matrix}
   p & q & r \\
   s & t & u \\
   v & w & x \\
\end{matrix} \right|=p\left| \begin{matrix}
   t & u \\
   w & x \\
\end{matrix} \right|-q\left| \begin{matrix}
   s & u \\
   v & x \\
\end{matrix} \right|+r\left| \begin{matrix}
   s & t \\
   v & w \\
\end{matrix} \right|$ and $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$. We then use these values to find the required values of x, y and z.

Complete step by step answer:
According to the problem, we are asked to solve the given system of linear equation by Crammer’s rule:
$x+y+z=1$
$ax+by+cz=k$
${{a}^{2}}x+{{b}^{2}}y+{{c}^{2}}z={{k}^{2}}$.
Let us write the given system of linear equations in the form of $AX=B$, where
A = coefficient matrix = $\left[ \begin{matrix}
   1 & 1 & 1 \\
   a & b & c \\
   {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\
\end{matrix} \right]$.
X = variable matrix = $\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]$.
B = constant matrix = $\left[ \begin{matrix}
   1 \\
   k \\
   {{k}^{2}} \\
\end{matrix} \right]$.
From cramer's rule, we know that the solution for system of linear equations is defined as $x=\dfrac{\left| {{A}_{1}} \right|}{\left| A \right|}$, $y=\dfrac{\left| {{A}_{2}} \right|}{\left| A \right|}$, $z=\dfrac{\left| {{A}_{3}} \right|}{\left| A \right|}$.
Where ${{A}_{1}}=\left[ \begin{matrix}
   1 & 1 & 1 \\
   k & b & c \\
   {{k}^{2}} & {{b}^{2}} & {{c}^{2}} \\
\end{matrix} \right]$, ${{A}_{2}}=\left[ \begin{matrix}
   1 & 1 & 1 \\
   a & k & c \\
   {{a}^{2}} & {{k}^{2}} & {{c}^{2}} \\
\end{matrix} \right]$, ${{A}_{3}}=\left[ \begin{matrix}
   1 & 1 & 1 \\
   a & b & k \\
   {{a}^{2}} & {{b}^{2}} & {{k}^{2}} \\
\end{matrix} \right]$.
Let us first find the values of $\left| A \right|$, $\left| {{A}_{1}} \right|$, $\left| {{A}_{2}} \right|$, $\left| {{A}_{3}} \right|$.
So, we have $\left| A \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   a & b & c \\
   {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\
\end{matrix} \right|$.
Let us perform the operations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$.
$\Rightarrow \left| A \right|=\left| \begin{matrix}
   1 & 1-1 & 1-1 \\
   a & b-a & c-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
$\Rightarrow \left| A \right|=\left| \begin{matrix}
   1 & 0 & 0 \\
   a & b-a & c-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
   p & q & r \\
   s & t & u \\
   v & w & x \\
\end{matrix} \right|=p\left| \begin{matrix}
   t & u \\
   w & x \\
\end{matrix} \right|-q\left| \begin{matrix}
   s & u \\
   v & x \\
\end{matrix} \right|+r\left| \begin{matrix}
   s & t \\
   v & w \\
\end{matrix} \right|$.
$\Rightarrow \left| A \right|=1\left| \begin{matrix}
   b-a & c-a \\
   {{b}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|-0\left| \begin{matrix}
   a & c-a \\
   {{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|+0\left| \begin{matrix}
   a & b-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
\[\Rightarrow \left| A \right|=\left( \left( b-a \right)\times \left( {{c}^{2}}-{{a}^{2}} \right) \right)-\left( \left( {{b}^{2}}-{{a}^{2}} \right)\times \left( c-a \right) \right)\].
\[\Rightarrow \left| A \right|=\left( b-a \right)\times \left( c-a \right)\times \left( \left( c+a \right)-\left( b+a \right) \right)\].
\[\Rightarrow \left| A \right|=\left( b-a \right)\times \left( c-a \right)\times \left( c-b \right)\] ---(1).
Now, we have $\left| {{A}_{1}} \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   k & b & c \\
   {{k}^{2}} & {{b}^{2}} & {{c}^{2}} \\
\end{matrix} \right|$.
Let us perform the operations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$.
$\Rightarrow \left| {{A}_{1}} \right|=\left| \begin{matrix}
   1 & 1-1 & 1-1 \\
   k & b-k & c-k \\
   {{k}^{2}} & {{b}^{2}}-{{k}^{2}} & {{c}^{2}}-{{k}^{2}} \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{1}} \right|=\left| \begin{matrix}
   1 & 0 & 0 \\
   k & b-k & c-k \\
   {{k}^{2}} & {{b}^{2}}-{{k}^{2}} & {{c}^{2}}-{{k}^{2}} \\
\end{matrix} \right|$.
We know that determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
   p & q & r \\
   s & t & u \\
   v & w & x \\
\end{matrix} \right|=p\left| \begin{matrix}
   t & u \\
   w & x \\
\end{matrix} \right|-q\left| \begin{matrix}
   s & u \\
   v & x \\
\end{matrix} \right|+r\left| \begin{matrix}
   s & t \\
   v & w \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{1}} \right|=1\left| \begin{matrix}
   b-k & c-k \\
   {{b}^{2}}-{{k}^{2}} & {{c}^{2}}-{{k}^{2}} \\
\end{matrix} \right|-0\left| \begin{matrix}
   k & c-k \\
   {{k}^{2}} & {{c}^{2}}-{{k}^{2}} \\
\end{matrix} \right|+0\left| \begin{matrix}
   k & b-k \\
   {{k}^{2}} & {{b}^{2}}-{{k}^{2}} \\
\end{matrix} \right|$.
We know that $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
\[\Rightarrow \left| {{A}_{1}} \right|=\left( \left( b-k \right)\times \left( {{c}^{2}}-{{k}^{2}} \right) \right)-\left( \left( {{b}^{2}}-{{k}^{2}} \right)\times \left( c-k \right) \right)\].
\[\Rightarrow \left| {{A}_{1}} \right|=\left( b-k \right)\times \left( c-k \right)\times \left( \left( c+k \right)-\left( b+k \right) \right)\].
\[\Rightarrow \left| {{A}_{1}} \right|=\left( b-k \right)\times \left( c-k \right)\times \left( c-b \right)\] ---(2).
So, we have $\left| {{A}_{2}} \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   a & k & c \\
   {{a}^{2}} & {{k}^{2}} & {{c}^{2}} \\
\end{matrix} \right|$.
Let us perform the operations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$.
$\Rightarrow \left| {{A}_{2}} \right|=\left| \begin{matrix}
   1 & 1-1 & 1-1 \\
   a & k-a & c-a \\
   {{a}^{2}} & {{k}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{2}} \right|=\left| \begin{matrix}
   1 & 0 & 0 \\
   a & k-a & c-a \\
   {{a}^{2}} & {{k}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
   p & q & r \\
   s & t & u \\
   v & w & x \\
\end{matrix} \right|=p\left| \begin{matrix}
   t & u \\
   w & x \\
\end{matrix} \right|-q\left| \begin{matrix}
   s & u \\
   v & x \\
\end{matrix} \right|+r\left| \begin{matrix}
   s & t \\
   v & w \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{2}} \right|=1\left| \begin{matrix}
   k-a & c-a \\
   {{k}^{2}}-{{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|-0\left| \begin{matrix}
   a & c-a \\
   {{a}^{2}} & {{c}^{2}}-{{a}^{2}} \\
\end{matrix} \right|+0\left| \begin{matrix}
   a & k-a \\
   {{a}^{2}} & {{k}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
\[\Rightarrow \left| {{A}_{2}} \right|=\left( \left( k-a \right)\times \left( {{c}^{2}}-{{a}^{2}} \right) \right)-\left( \left( {{k}^{2}}-{{a}^{2}} \right)\times \left( c-a \right) \right)\].
\[\Rightarrow \left| {{A}_{2}} \right|=\left( k-a \right)\times \left( c-a \right)\times \left( \left( c+a \right)-\left( k+a \right) \right)\].
\[\Rightarrow \left| {{A}_{2}} \right|=\left( k-a \right)\times \left( c-a \right)\times \left( c-k \right)\] ---(3).
So, we have $\left| {{A}_{3}} \right|=\left| \begin{matrix}
   1 & 1 & 1 \\
   a & b & k \\
   {{a}^{2}} & {{b}^{2}} & {{k}^{2}} \\
\end{matrix} \right|$.
Let us perform the operations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$.
$\Rightarrow \left| {{A}_{3}} \right|=\left| \begin{matrix}
   1 & 1-1 & 1-1 \\
   a & b-a & k-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} & {{k}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{3}} \right|=\left| \begin{matrix}
   1 & 0 & 0 \\
   a & b-a & k-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} & {{k}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
   p & q & r \\
   s & t & u \\
   v & w & x \\
\end{matrix} \right|=p\left| \begin{matrix}
   t & u \\
   w & x \\
\end{matrix} \right|-q\left| \begin{matrix}
   s & u \\
   v & x \\
\end{matrix} \right|+r\left| \begin{matrix}
   s & t \\
   v & w \\
\end{matrix} \right|$.
$\Rightarrow \left| {{A}_{3}} \right|=1\left| \begin{matrix}
   b-a & k-a \\
   {{b}^{2}}-{{a}^{2}} & {{k}^{2}}-{{a}^{2}} \\
\end{matrix} \right|-0\left| \begin{matrix}
   a & k-a \\
   {{a}^{2}} & {{k}^{2}}-{{a}^{2}} \\
\end{matrix} \right|+0\left| \begin{matrix}
   a & b-a \\
   {{a}^{2}} & {{b}^{2}}-{{a}^{2}} \\
\end{matrix} \right|$.
We know that $\left| \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
\[\Rightarrow \left| {{A}_{3}} \right|=\left( \left( b-a \right)\times \left( {{k}^{2}}-{{a}^{2}} \right) \right)-\left( \left( {{b}^{2}}-{{a}^{2}} \right)\times \left( k-a \right) \right)\].
\[\Rightarrow \left| {{A}_{3}} \right|=\left( b-a \right)\times \left( k-a \right)\times \left( \left( k+a \right)-\left( b+a \right) \right)\].
\[\Rightarrow \left| {{A}_{3}} \right|=\left( b-a \right)\times \left( k-a \right)\times \left( k-b \right)\] ---(4).
Now, let us find the values of x, y and z.
So, we have $x=\dfrac{\left| A \right|}{\left| {{A}_{1}} \right|}$.
$\Rightarrow x=\dfrac{\left( b-a \right)\times \left( c-a \right)\times \left( c-b \right)}{\left( b-k \right)\times \left( c-k \right)\times \left( c-b \right)}$.
$\Rightarrow x=\dfrac{\left( b-a \right)\left( c-a \right)}{\left( b-k \right)\left( c-k \right)}$.
Now, we have $y=\dfrac{\left| A \right|}{\left| {{A}_{2}} \right|}$.
$\Rightarrow y=\dfrac{\left( b-a \right)\times \left( c-a \right)\times \left( c-b \right)}{\left( k-a \right)\times \left( c-a \right)\times \left( c-k \right)}$.
$\Rightarrow y=\dfrac{\left( b-a \right)\left( c-b \right)}{\left( k-a \right)\left( c-k \right)}$.
Now, we have $z=\dfrac{\left| A \right|}{\left| {{A}_{3}} \right|}$.
$\Rightarrow z=\dfrac{\left( b-a \right)\times \left( c-a \right)\times \left( c-b \right)}{\left( b-a \right)\times \left( k-a \right)\times \left( k-b \right)}$.
$\Rightarrow z=\dfrac{\left( c-a \right)\left( c-b \right)}{\left( k-a \right)\left( k-b \right)}$.

$\therefore $ We have found the solution set of the given linear equations as $x=\dfrac{\left( b-a \right)\left( c-a \right)}{\left( b-k \right)\left( c-k \right)}$, $y=\dfrac{\left( b-a \right)\left( c-b \right)}{\left( k-a \right)\left( c-k \right)}$, $z=\dfrac{\left( c-a \right)\left( c-b \right)}{\left( k-a \right)\left( k-b \right)}$.

Note: We can see that the given problem contains a huge amount of calculations so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We should not confuse while writing the matrices ${{A}_{1}}$, ${{A}_{2}}$, ${{A}_{3}}$ as writing these matrices wrongly will lead us to incorrect solutions. Similarly, we can expect the problems to find the solution of the given system of linear equations by using matrix inversion method or Gauss-Jordan method.