Browse Questions

# Using matrices, solve the following system of equations : $x+2y-3z=-4, 2x+3y+2z=2 \: and \: 3x-3y-4z=11$

Toolbox:
• Minor of an element $a_{ij}$ of a determinant is the determinant obtained by deleting it $i^{th}$ row and $j^{th}$ column in which $a_{ij}$ lies.Minor of an element $a_{ij}$ is denoted by $M_{ij}$
• Cofactor of an element $a_{ij}$ denoted by $A_{ij}$ is defined by $A_{ij}=(-1)^{i+j}M_{ij}$,where $M_{ij}$ is minor of $a_{ij}$
Step 1:
Given :
$x+2y-3z=-4,2x+3y+2z=2,3x-3y-4z=11$
This can be written in the matrix form as
$\begin{bmatrix}1 & 2& 3\\2 & 3& 2\\3 &-3 &-4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-4\\2\\11\end{bmatrix}$
(i.e) $Ax=B$
Here $A=\begin{bmatrix}1 &2&-3\\2 &3&2\\3 &-3 &-4\end{bmatrix}$
$\mid A\mid=\begin{bmatrix}1 & 2 & -3\\2 & 3&2\\3 &-3&-4\end{bmatrix}$
$\qquad=-6+28+45=67\neq 0$
Hence $A$ is invertible.
Step 2:
Now let us find the adj of A
$C_{11}=(-1)^{1+1}\begin{vmatrix}3 &2\\-3 & -4\end{vmatrix}=-6$
$C_{12}=(-1)^{1+2}\begin{vmatrix}2 &2\\3 & -4\end{vmatrix}=14$
$C_{13}=(-1)^{1+3}\begin{vmatrix}2&3\\3 & -3\end{vmatrix}=-15$
$C_{21}=(-1)^{2+1}\begin{vmatrix}2 &-3\\-3 & -4\end{vmatrix}=17$
$C_{22}=(-1)^{2+2}\begin{vmatrix}1 &-3\\3 & -4\end{vmatrix}=5$
$C_{32}=(-1)^{3+2}\begin{vmatrix}1 &-3\\2 & 2\end{vmatrix}=-8$
$C_{33}=(-1)^{3+3}\begin{vmatrix}1 &2\\2 & 3\end{vmatrix}=-1$
Step 3:
Hence adj of $A$ is
$\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}=\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}$
$A^{-1}=\large\frac{1}{\mid A\mid}$$adj A \quad=\large\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix} Step 4: X=A^{-1}B \begin{bmatrix}x\\y\\z\end{bmatrix}=\large\frac{1}{67}\begin{bmatrix}-6&17&13\\14&5&-8\\-15&9&-1\end{bmatrix}\begin{bmatrix}4\\2\\11\end{bmatrix} \Rightarrow \large\frac{1}{67}$$\begin{bmatrix}24+34+143\\-56+10-88\\60+18-11\end{bmatrix}$
$\Rightarrow \begin{bmatrix}3\\2\\1\end{bmatrix}$
Hence $x=3,y=2$ and $z=1$ are the required solution.