Answer
414.6k+ views
Hint: The inverse of the square matrix exists, where the determinant of the matrix does not equal to zero i.e. the non-singular. It follows the property - ${{A}^{-1}}A={{A}^{-1}}A=I$ where I is the identity matrix. By using elementary column transformation, you swap (interchanged) the columns and multiply the column with non-zero constant and add a multiple of one column to another column.
Complete step-by-step answer:
To find the inverse, first find the determinant of A
Determinant is an element that determines or identifies the nature or conditions of an outcome. It is denoted by det(A), det A or $\left| A \right|$
$\Rightarrow \left| A \right|=\left| \begin{matrix}
1 & 0 & 3 \\
0 & 2 & 3 \\
1 & 2 & 1 \\
\end{matrix} \right|$
The determinant of A is the product of the diagonal entries of the row echelon and the times of a factor $\pm 1$.
Now,
$\Rightarrow \left| A \right|=1[(2)(1)-(2)(3)]-0[(0)(1)-(1)(3)]+1[(0)(2)-(2)(1)]$
Simplify the left hand side of the equation –
Here, the middle term is equal to zero, as zero multiplied with anything becomes zero.
$\begin{align}
\Rightarrow & \left| A \right|=1[2-6]-0+1[-2] \\
& \left| A \right|=-4-2 \\
\end{align}$
Here, both the terms are negative so according to the property, minus and minus do plus and sign of minus.
$\begin{align}
\Rightarrow & \left| A \right|=-6 \\
\Rightarrow & \left| A \right|\ne 0 \\
\end{align}$
The determinant is not equal to zero, therefore, the inverse exists.
Now, consider ${{A}^{-1}}A=I$
Place, the values of the given matrix A in the above equation
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]$
Now, subtract column one from column third ${{C}_{3}}-{{C}_{1}}$
We get,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 2 & 3 \\
1 & 2 & 0 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]$
Now interchange column second and third,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 3 & 2 \\
1 & 0 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{matrix} \right]$
Now, subtract column third from second –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 2 \\
1 & -2 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 0 \\
0 & -1 & 1 \\
0 & 1 & 0 \\
\end{matrix} \right]$
Now, multiply column second with two and subtract it from column third –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & -2 & 6 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 2 \\
0 & -1 & 3 \\
0 & 1 & -2 \\
\end{matrix} \right]$
Divide column third by six, $(\dfrac{{{C}_{3}}}{6})$
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & -2 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & \dfrac{1}{3} \\
0 & -1 & \dfrac{1}{2} \\
0 & 1 & \dfrac{1}{-3} \\
\end{matrix} \right]$
Now, use ${{C}_{1}}-{{C}_{3}}\text{ and }{{\text{C}}_{2}}+2{{C}_{3}}$ operations –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
\dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\
\dfrac{-1}{2} & 0 & \dfrac{1}{2} \\
\dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\
\end{matrix} \right]$
Now, take $\dfrac{1}{6}$ common from the right hand side of the equation –
${{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix}
4 & -2 & 2 \\
-3 & 0 & 3 \\
2 & 2 & -2 \\
\end{matrix} \right]$
This is the required solution.
Note: Always read the given instructions twice, as the inverse of any matrix can be calculated by the row and column transformation. Also it can be calculated by the adjoint method. So, apply accordingly.
Complete step-by-step answer:
To find the inverse, first find the determinant of A
Determinant is an element that determines or identifies the nature or conditions of an outcome. It is denoted by det(A), det A or $\left| A \right|$
$\Rightarrow \left| A \right|=\left| \begin{matrix}
1 & 0 & 3 \\
0 & 2 & 3 \\
1 & 2 & 1 \\
\end{matrix} \right|$
The determinant of A is the product of the diagonal entries of the row echelon and the times of a factor $\pm 1$.
Now,
$\Rightarrow \left| A \right|=1[(2)(1)-(2)(3)]-0[(0)(1)-(1)(3)]+1[(0)(2)-(2)(1)]$
Simplify the left hand side of the equation –
Here, the middle term is equal to zero, as zero multiplied with anything becomes zero.
$\begin{align}
\Rightarrow & \left| A \right|=1[2-6]-0+1[-2] \\
& \left| A \right|=-4-2 \\
\end{align}$
Here, both the terms are negative so according to the property, minus and minus do plus and sign of minus.
$\begin{align}
\Rightarrow & \left| A \right|=-6 \\
\Rightarrow & \left| A \right|\ne 0 \\
\end{align}$
The determinant is not equal to zero, therefore, the inverse exists.
Now, consider ${{A}^{-1}}A=I$
Place, the values of the given matrix A in the above equation
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]$
Now, subtract column one from column third ${{C}_{3}}-{{C}_{1}}$
We get,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 2 & 3 \\
1 & 2 & 0 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]$
Now interchange column second and third,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 3 & 2 \\
1 & 0 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{matrix} \right]$
Now, subtract column third from second –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 2 \\
1 & -2 & 2 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 0 \\
0 & -1 & 1 \\
0 & 1 & 0 \\
\end{matrix} \right]$
Now, multiply column second with two and subtract it from column third –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & -2 & 6 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & 2 \\
0 & -1 & 3 \\
0 & 1 & -2 \\
\end{matrix} \right]$
Divide column third by six, $(\dfrac{{{C}_{3}}}{6})$
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & -2 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 & -1 & \dfrac{1}{3} \\
0 & -1 & \dfrac{1}{2} \\
0 & 1 & \dfrac{1}{-3} \\
\end{matrix} \right]$
Now, use ${{C}_{1}}-{{C}_{3}}\text{ and }{{\text{C}}_{2}}+2{{C}_{3}}$ operations –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
\dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\
\dfrac{-1}{2} & 0 & \dfrac{1}{2} \\
\dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\
\end{matrix} \right]$
Now, take $\dfrac{1}{6}$ common from the right hand side of the equation –
${{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix}
4 & -2 & 2 \\
-3 & 0 & 3 \\
2 & 2 & -2 \\
\end{matrix} \right]$
This is the required solution.
Note: Always read the given instructions twice, as the inverse of any matrix can be calculated by the row and column transformation. Also it can be calculated by the adjoint method. So, apply accordingly.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)