Question

The sum of three numbers is $6$. If we multiply the third number by $3$ and add it to the second number we get $11$. By adding first and third numbers we get a number, which is double than the second number. Use this information and find a system of linear equations. Find these three numbers using matrices.

Answer 1

Let the first , second & third number be $x,y,z$ respectively
Given,
$\therefore x+y+z=6$
$y+3z=11$
$x+z=2y$ or $x-2y+z=0$
Step 1
Write equation as $AX=B$
$A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]$
Hence $A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ & $B=\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]$
Step 2
Calculate $|A|$
$|A|=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right]$
$=1(1+6)-0(1+2)+1(3+1)$
$=7+2$
$=9$
SO , $|A|\ne 0$
$\therefore$ The system of equation is consistent & has a unique solutions
Now $AX=B$
$X=A^{-1}B$
Hence $A=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ & $B=\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]$
$=1(1+6)-0(1+2)+1(3-1)$
$=7+2$
$=9\ne 0$
Since determinant is not equal to O, $A^{-1}$ exists.
Now find adj (A)
Now $A=B$
$X=A^{-1}B$
Step 3
Calculating $X=A^{-1}B$
Calculating $A^{-1}$
Now $A^{-1}=\frac{1}{|A|}ajd\left(A\right)$
adj A =
${\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^{\prime }=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]$
$A=\left[\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right]$
$A_{11}=1\times 1-3\times \left(2\right)=1+6=7$
$A_{12}=-[0\times 1-3\times 1]=-(-3)=3$
$A_{13}=-0\times (-2)-1\times 1=-1$
$A_{21}=[1\times 1-(-2)\times 1]=-[1+2]=-3$
$A_{22}=1\times 1-1\times 1=1-1=0$
$A_{23}=[1\times (-2)-1\times 1]=-[-2-1]=-(-3)=3$
$A_{31}=1\times 3-1\times 1=3-1=2$
$A_{32}=-[1\times 3-0\times 1]=-[3-0]=-3$
$A_{33}=1\times 1-1\times 0=1-0=1$
Hence, $adj\left(A\right)=\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]$
Now ,
$A^{-1}=\frac{1}{|A|}adj\left(A\right)$
$A^{-1}=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]$
Solution of given system of equations is
$X=A^{-1}B$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]\left[\begin{array}{c}6\\ 11\\ 0\end{array}\right]$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{ccc}42& -33& +0\\ 18& +0& +0\\ -6& +33& +0\end{array}\right]$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}9\\ 18\\ 27\end{array}\right]$
$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$
$\therefore x=1,y=2,z=3$