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Question: Find the inverse of the matrix \(\left[ \begin{matrix} 1 & 0 & 1 \\\ 0 & 2 & 3 \\\ 1 ...

Find the inverse of the matrix [101 023 121 ]\left[ \begin{matrix} 1 & 0 & 1 \\\ 0 & 2 & 3 \\\ 1 & 2 & 1 \\\ \end{matrix} \right] by using elementary column transformation.

Explanation

Solution

To solve the question given above, we will first find out what is a matrix. Then we will find out what are elementary column transformations and what are the kind of elementary transformations we can apply. Then, we will assume that the given matrix is A. We will write A = AI. Using appropriate elementary column transformations, we will write the above equation as I = AB where B will be the inverse of matrix A.

Complete step by step solution:
Before we start to solve the question given above, we must know what is a matrix. A matrix is a rectangular array or table of numbers, or expressions, arranged in rows and columns. Elementary column transformations are the operations performed on columns of the matrices to transform it into a different form so that the calculations become simpler. We can only apply following elementary column transformations:
(i) We can interchange the column within the same matrix.
(ii) We can multiply the entire column with a non-zero number.
(iii) We can add one column to another column multiplied by a non-zero number.
Now, let us assume that the matrix given in question is A. Thus, we have:
A=[101 023 121 ]A=\left[ \begin{matrix} 1 & 0 & 1 \\\ 0 & 2 & 3 \\\ 1 & 2 & 1 \\\ \end{matrix} \right]
Now, we can write A also as:
A=IA ⇒[101 023 121 ]=[100 010 001 ]A \begin{aligned} & A=IA \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 1 \\\ 0 & 2 & 3 \\\ 1 & 2 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]A \\\ \end{aligned}
Now, we will subtract the first column from third column. Thus, we will get:
⇒[10(1−1) 02(3−0) 12(1−1) ]=[10(0−1) 01(0−0) 00(1−0) ]A ⇒[100 023 120 ]=[10−1 010 001 ]A \begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & \left( 1-1 \right) \\\ 0 & 2 & \left( 3-0 \right) \\\ 1 & 2 & \left( 1-1 \right) \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \left( 0-1 \right) \\\ 0 & 1 & \left( 0-0 \right) \\\ 0 & 0 & \left( 1-0 \right) \\\ \end{matrix} \right]A \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 2 & 3 \\\ 1 & 2 & 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & -1 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]A \\\ \end{aligned}
Now, we will subtract the second column multiplied by (32)\left( \dfrac{3}{2} \right) from third column. Thus, we will get:
⇒[10(0−32×0) 02(3−32(2)) 12(0−32(2)) ]=[10(−1−32×0) 01(0−32(1)) 00(1−32×0) ]A ⇒[100 020 12−3 ]=[10−1 01−32 001 ]A \begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & \left( 0-\dfrac{3}{2}\times 0 \right) \\\ 0 & 2 & \left( 3-\dfrac{3}{2}\left( 2 \right) \right) \\\ 1 & 2 & \left( 0-\dfrac{3}{2}\left( 2 \right) \right) \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & \left( -1-\dfrac{3}{2}\times 0 \right) \\\ 0 & 1 & \left( 0-\dfrac{3}{2}\left( 1 \right) \right) \\\ 0 & 0 & \left( 1-\dfrac{3}{2}\times 0 \right) \\\ \end{matrix} \right]A \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 2 & 0 \\\ 1 & 2 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & -1 \\\ 0 & 1 & \dfrac{-3}{2} \\\ 0 & 0 & 1 \\\ \end{matrix} \right]A \\\ \end{aligned}
Now, we will add the third column multiplied by (23)\left( \dfrac{2}{3} \right) to second column. Thus, we will get:
⇒[1(0+23(0))0 0(2+23(0))0 1(2+23(−3))−3 ]=[10+23(−1)−1 01+23(−32)−32 00+23(1)1 ]A ⇒[100 020 10−3 ]=[1−23−1 00−32 0231 ]A \begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & \left( 0+\dfrac{2}{3}\left( 0 \right) \right) & 0 \\\ 0 & \left( 2+\dfrac{2}{3}\left( 0 \right) \right) & 0 \\\ 1 & \left( 2+\dfrac{2}{3}\left( -3 \right) \right) & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0+\dfrac{2}{3}\left( -1 \right) & -1 \\\ 0 & 1+\dfrac{2}{3}\left( \dfrac{-3}{2} \right) & \dfrac{-3}{2} \\\ 0 & 0+\dfrac{2}{3}\left( 1 \right) & 1 \\\ \end{matrix} \right]A \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 2 & 0 \\\ 1 & 0 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & \dfrac{-2}{3} & -1 \\\ 0 & 0 & \dfrac{-3}{2} \\\ 0 & \dfrac{2}{3} & 1 \\\ \end{matrix} \right]A \\\ \end{aligned}
Now, we will add the third column multiplied by (13)\left( \dfrac{1}{3} \right) to first column. Thus, we will get:
⇒[1+(13×0)00 0+(13×0)20 1+(13×(−1))0−3 ]=[1+13(−1)−23−1 0+13(−32)0−32 0+13(1)231 ]A ⇒[100 020 00−3 ]=[23−23−1 −120−32 13231 ]A \begin{aligned} & \Rightarrow \left[ \begin{matrix} 1+\left( \dfrac{1}{3}\times 0 \right) & 0 & 0 \\\ 0+\left( \dfrac{1}{3}\times 0 \right) & 2 & 0 \\\ 1+\left( \dfrac{1}{3}\times \left( -1 \right) \right) & 0 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1+\dfrac{1}{3}\left( -1 \right) & \dfrac{-2}{3} & -1 \\\ 0+\dfrac{1}{3}\left( \dfrac{-3}{2} \right) & 0 & \dfrac{-3}{2} \\\ 0+\dfrac{1}{3}\left( 1 \right) & \dfrac{2}{3} & 1 \\\ \end{matrix} \right]A \\\ & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 2 & 0 \\\ 0 & 0 & -3 \\\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{2}{3} & \dfrac{-2}{3} & -1 \\\ \dfrac{-1}{2} & 0 & \dfrac{-3}{2} \\\ \dfrac{1}{3} & \dfrac{2}{3} & 1 \\\ \end{matrix} \right]A \\\ \end{aligned}
Now, we will divide the second column by 2 and third column by (-3). Thus, we will get:
⇒[100 010 001 ]=[23−1313 −12012 1313−13 ]A ⇒I=[23−1313 −12012 1313−13 ]A............(1) \begin{aligned} & \Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\\ \dfrac{-1}{2} & 0 & \dfrac{1}{2} \\\ \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\\ \end{matrix} \right]A \\\ & \Rightarrow I=\left[ \begin{matrix} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\\ \dfrac{-1}{2} & 0 & \dfrac{1}{2} \\\ \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\\ \end{matrix} \right]A............\left( 1 \right) \\\ \end{aligned}
We know that:
I=AA−1..................(2)I=A{{A}^{-1}}..................\left( 2 \right)
On comparing (1) and (2), we will get:
A−1=[23−1313 −12012 1313−13 ]{{A}^{-1}}=\left[ \begin{matrix} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\\ \dfrac{-1}{2} & 0 & \dfrac{1}{2} \\\ \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\\ \end{matrix} \right]
Thus, the above matrix is the required inverse matrix.

Note: One important thing to remember is that the inverse does not exist for any matrix. Inverse only exists for those matrices whose determinant is non-zero. Thus, if a matrix is given and if we have to find its inverse matrix and its determinant is zero, then we will not be able to find its inverse.