Question

Solve by matrix inversion method: 2x – y = 7; 3x – 2y = 11.

Answer 1

Hint: First write a list of all the co – efficients of variables. Now fill a \[2\times 2\] matrix in order with the terms from the list. Now make a list of constants. Make a matrix from them. Now make a list of variables. Make a matrix from them. Now write a product of matrices from co – efficient and variables. Find the product, equate this product to a constant matrix. You must set both equations. From this you get the equation of the system. Now find the inverse of the matrix from co – efficients. Now multiply this with the matrix of constants. You get the solutions.

Complete step-by-step solution -
Matrix Inversion method: - When you have 2 equations \[ax+by=c\], \[dx+ey=f\] we can write this system as:
\[\left[ \begin{matrix}
   a & b \\
   d & e \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   c \\
   f \\
\end{matrix} \right]\]
Find the inverse of the first matrix and multiply of both sides. We get,
\[{{A}^{-1}}A\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]={{A}^{-1}}\left[ \begin{matrix}
   c \\
   f \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]={{A}^{-1}}\left[ \begin{matrix}
   c \\
   f \\
\end{matrix} \right]\], this gives us the solutions.
Here we are given system of equations, by the 2 equations:
2x – y = 7 – (i); 3x – 2y = 11 – (ii)
The list of all the co – efficients are given by {2, -1, 3, -2}.
By forming a matric from the above list, we got it as: \[\left[ \begin{matrix}
   2 & -1 \\
   3 & -2 \\
\end{matrix} \right]\].
The list of all the variables is given by {x, y}.
By forming a matric from the above list, we get it as: \[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]\].
The list of all the constants is given by {7, 11}.
By forming a matrix from the above list, we get it as: \[\left[ \begin{matrix}
   7 \\
   11 \\
\end{matrix} \right]\].
Now, we need an inverse of co – efficient matrix.
In \[2\times 2\] matrix inverse of \[A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\] is given by:
\[{{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}.....\left( 1 \right)\]
Adjoint of a matrix is the transpose of a cofactor matrix.
Cofactor matrix is found by replacing each and every term with its cofactor.
A cofactor is the number you get when you remove the column and row of a designated element in a matrix, which is just a numerical grid in the form of a rectangle or a square. The cofactor is always preceded by a positive (+) or negative (-) sign, depending whether the element is in a + or - position.
By using above definition, we can write it as:
Cofactor of the term ‘a’ is the term ‘d’.
Cofactor of the term ‘b’ is the term ‘-c’.
Cofactor of the term ‘c’ is the term ‘-b’.
Cofactor of the term ‘d’ is the term ‘a’.
By these we can say cofactor matrix of A as:
\[\text{Cofactor matrix of }A=\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]\]
We know the relation between adjoint of matrix and cofactor matrix as:
\[Adj\left( A \right)={{\left( Cofactor\text{ }matrix \right)}^{T}}\]
By using above equation, we can write it as:
\[\text{ }Adj\left( A \right)={{\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   d & -b \\
   -c & a \\
\end{matrix} \right]\]
 By using determinant formula, we can write the det as:
\[\det A=\left( a \right)\left( d \right)-\left( b \right)\left( c \right)=ad-bc\]
By substituting values of det and adj into equation (1), we get the inverse of A as:
\[{{A}^{-1}}=\dfrac{1}{ad-bc}\left[ \begin{matrix}
   d & -b \\
   -c & a \\
\end{matrix} \right]\]
So, here we have a = 2, b = -1, c = 3, d = -2. So, we get:
\[{{A}^{-1}}=\dfrac{1}{-1}\left[ \begin{matrix}
   -2 & 1 \\
   -3 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 & -1 \\
   3 & -2 \\
\end{matrix} \right]\]
So, by matrix inversion method, we get the equation,
\[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]={{A}^{-1}}\left[ \begin{matrix}
   7 \\
   11 \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 & -1 \\
   3 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   7 \\
   11 \\
\end{matrix} \right]\]
So, multiplying the first row, with column we get,
2 (7) – 11 = 14 – 11 = 3
The above term will be first term of the matrix.
By multiplying second row with column, we get:
3 (7) – 2(11) = 21 – 22 = -1
The above term will be the second term of the matrix.
By substituting these both into one equation, we get:
\[\left[ \begin{matrix}
   x \\
   y \\
\end{matrix} \right]=\left[ \begin{matrix}
   3 \\
   -1 \\
\end{matrix} \right]\]
By substituting the values x = 3, y = -1, in equation (i), we get:
2 (3) – (- 1) = 7
7 = 7
Hence verified
So, this is true. So, one solution is (3, -1).

Note: Be careful while finding the inverse, because each and every sign matters. Multiply the inverse with the constant on the left side only if you put the inverse on the right side you will get the wrong answer, because in matrices \[AB\ne BA\] sometimes. So, this may affect your answer. In these types of solutions the verification must step to check your result.