Using matrix method, solve the following system of equations: l x-2 y+3 z=6 x+4 y+z=12 x-3 y+2 z=1

The given system of equations can be written as AX = B where A= [ rrr 1 & -2 & 3 1 & 4 & 1 1 & -3 & 2 ], X= [ l x y z ] and B= [ r 6 12 1 ]. Now, | A |=1(8+3)+2(2-1)+3(-3-4)=-8 0 A^ -1 exists the given system of equations has a unique solution X=A^ -1 B l A_ 11 =11, A_ 12 =-1, A_ 13 =-7, A_ 21 =-5, A_ 22 =-1, A_ 23 =1, A_ 31 =-14, A_ 32 =2, A_ 33 =6. l So, A^ -1 =1 -8 [ rrr 11 & -5 & -14 -1 & -1 & 2 -7 & 1 & 6 ] [ l x y z ]=1 -8 [ rrr 11 & -5 & -14 -1 & -1 & 2 -7 & 1 & 6 ] [ r 6 12 1 ] =1 -8 [ r 66-60-14 -6-12+2 -42+12+6 ]=1 -8 [ r -8 -16 -24 ] x=1, y=2, z =3

Using matrix method, solve the following system of equations: l x…

Download App

Question

Using matrix method, solve the following system of equations:
$
\begin{array}{l}
x-2 y+3 z=6 \\
x+4 y+z=12 \\
x-3 y+2 z=1
\end{array}
$

✓

Answer

The given system of equations can be written as $AX = B$
$
\text { where } A=\left[\begin{array}{rrr}
1 & -2 & 3 \\
1 & 4 & 1 \\
1 & -3 & 2
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{r}
6 \\
12 \\
1
\end{array}\right].
$
Now, $| A |=1(8+3)+2(2-1)+3(-3-4)=-8 \neq 0$
$
\Rightarrow A^{-1} \text { exists }
$
$\Rightarrow$ the given system of equations has a unique solution $X=A^{-1} B$
$
\begin{array}{l}
A_{11}=11, A_{12}=-1, A_{13}=-7, \\
A_{21}=-5, A_{22}=-1, A_{23}=1, \\
A_{31}=-14, A_{32}=2, A_{33}=6.
\end{array}
$
$\begin{array}{l}\text { So, } A^{-1}=\frac{1}{-8}\left[\begin{array}{rrr}11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6\end{array}\right] \\ \therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-8}\left[\begin{array}{rrr}11 & -5 & -14 \\ -1 & -1 & 2 \\ -7 & 1 & 6\end{array}\right]\left[\begin{array}{r}6 \\ 12 \\ 1\end{array}\right] \\ =\frac{1}{-8}\left[\begin{array}{r}66-60-14 \\ -6-12+2 \\ -42+12+6\end{array}\right]=\frac{1}{-8}\left[\begin{array}{r}-8 \\ -16 \\ -24\end{array}\right] \\ \Rightarrow x=1, y=2, z =3\end{array}$