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First, let me show you some notation.
To show a matrix is an inverse of another matrix, we write
-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.
For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.
Then multiply each entry in the matrix by 1 divided by the determinant.
![\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] = \\ \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%20%3D%20%0A%20%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26%7B-b%7D%5C%5C%7B-c%7D%26a%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5C%5C%20%20%5C%5C%20%5C%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%26%7B-b%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5C%5C%20%7B-c%7D%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%26a%28%5Cfrac%7B1%7D%7Bad-bc%7D%29%20%5Cend%7Barray%7D%5Cright%5D)
I hope this helped!
To show a matrix is an inverse of another matrix, we write
-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.
For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.
Then multiply each entry in the matrix by 1 divided by the determinant.
I hope this helped!
Answer:
A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.
The given equation consisting of variable , m is
None of the two solution
, is extraneous.
Here, L.H.S= R.H.S
Option A: 0→ extraneous