
For any non singular matrix A, \[{{A}^{-1}}=\]
A.\[\left| A \right|adj\left( A \right)\]
B.\[\dfrac{1}{\left| A \right|adj\left( A \right)}\]
C.\[\dfrac{adj\left( A \right)}{\left| A \right|}\]
D.None of these
Answer
596.4k+ views
Hint: -In mathematics, a matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. For example, the dimension of the matrix below is \[2\times 3\] , because there are two rows and three columns.
\[\left[ \begin{matrix}
1 & 9 & -13 \\
20 & 5 & -6 \\
\end{matrix} \right]\]
Complete step-by-step answer:
Provided that they have the same size (each matrix has the same number of rows and the same number of columns as the other), two matrices can be added or subtracted element by element.
A non-singular matrix is a matrix whose determinant is non-zero.
Inverse of a matrix can be calculated or is defined only if the matrix is non-singular and is a square matrix which means its rows and columns are equal.
Other facts that are useful to find the solution of the question are as follows
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
(Where n is the order of the matrix A)
\[A\cdot {{A}^{-1}}={{I}_{n}}\]
(Where n is the order of the matrix A)
As mentioned in the question, we have to find the value of \[{{A}^{-1}}\] .
Now, as mentioned in the hint, we get
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
Now, we can write the above expression as follows
\[\dfrac{A\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\]
Now, on multiplying \[{{A}^{-1}}\] to both the sides of the equation, we get
\[\dfrac{\left( {{A}^{-1}} \right)\cdot A\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\cdot \left( {{A}^{-1}} \right)\]
Now, as mentioned in the hint, we get
\[\begin{align}
& \dfrac{{{I}_{n}}\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\cdot \left( {{A}^{-1}} \right) \\
& \dfrac{adj\left( A \right)}{\left| A \right|}={{A}^{-1}} \\
\end{align}\]
Hence, the correct option in (c).
Note: -The students can make an error if they don’t know about the properties of a matrix that are given in the hint as follows
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
(Where n is the order of the matrix A)
\[A\cdot {{A}^{-1}}={{I}_{n}}\]
(Where n is the order of the matrix A)
Without knowing these above mentioned properties, one cannot get to the right answer.
\[\left[ \begin{matrix}
1 & 9 & -13 \\
20 & 5 & -6 \\
\end{matrix} \right]\]
Complete step-by-step answer:
Provided that they have the same size (each matrix has the same number of rows and the same number of columns as the other), two matrices can be added or subtracted element by element.
A non-singular matrix is a matrix whose determinant is non-zero.
Inverse of a matrix can be calculated or is defined only if the matrix is non-singular and is a square matrix which means its rows and columns are equal.
Other facts that are useful to find the solution of the question are as follows
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
(Where n is the order of the matrix A)
\[A\cdot {{A}^{-1}}={{I}_{n}}\]
(Where n is the order of the matrix A)
As mentioned in the question, we have to find the value of \[{{A}^{-1}}\] .
Now, as mentioned in the hint, we get
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
Now, we can write the above expression as follows
\[\dfrac{A\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\]
Now, on multiplying \[{{A}^{-1}}\] to both the sides of the equation, we get
\[\dfrac{\left( {{A}^{-1}} \right)\cdot A\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\cdot \left( {{A}^{-1}} \right)\]
Now, as mentioned in the hint, we get
\[\begin{align}
& \dfrac{{{I}_{n}}\cdot adj\left( A \right)}{\left| A \right|}={{I}_{n}}\cdot \left( {{A}^{-1}} \right) \\
& \dfrac{adj\left( A \right)}{\left| A \right|}={{A}^{-1}} \\
\end{align}\]
Hence, the correct option in (c).
Note: -The students can make an error if they don’t know about the properties of a matrix that are given in the hint as follows
\[A\cdot adj\left( A \right)=\left| A \right|{{I}_{n}}\]
(Where n is the order of the matrix A)
\[A\cdot {{A}^{-1}}={{I}_{n}}\]
(Where n is the order of the matrix A)
Without knowing these above mentioned properties, one cannot get to the right answer.
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