What is the condition of a matrix to have an inverse? Find the inverse of the matrix \(A = \begin{bmatrix} 5 & 1 & 2\\ 1 & 0 & 3\\ 4 & -3 & 8 \end{bmatrix}\)  If it exists.

This answer is restricted. Please login to view the answer of this question.

Login Now

The condition of a matrix to have an inverse is If A is an invertible matrix then there is a matrix C such that

AC = I = CA

In such case, C is called inverse of A and write as C = A-1

Problem Part:

Given

A = \(\begin{bmatrix}1 & -2 & -1\\ -1 & 5 & 6\\ 5 & -4 & 5\end{bmatrix}\)

~ [A:I]

\(~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ -1 & 5 & 6 & : & 0 & 1 & 0\\ 5 & -4 & 5 & : & 0 & 0 & 1\end{bmatrix}\)

Applying Reduced Echelon Form

R2 → R2 + R1,   R3 → R3 – 5R1

\(~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ 0 & 3 & 5 & : & 1 & 1 & 0\\ 0 & 6 & 10 & : & -5 & 0 & 1\end{bmatrix}\)

\(R_2 \to \frac{1}{3} R_2\)

\(~\begin{bmatrix}1 & -2 & -1 & : & 1 & 0 & 0\\ 0 & 1 & \frac{5}{3} & : & \frac{1}{3} & \frac{1}{3} & 0\\ 0 & 6 & 10 & : & -5 & 0 & 1\end{bmatrix}\)

R1 → R1 + 2R2 ,   R3 → R3 – 6R2

\(~\begin{bmatrix}1 & 0 & \frac{7}{3} & : & \frac{5}{3} & \frac{2}{3} & 0\\ 0 & 1 & \frac{5}{3} & : & \frac{1}{3} & \frac{1}{3} & 0\\ 0 & 0 & 0 & : & -7 & -2 & 1\end{bmatrix}\)

To have inverse, We should have one pivot element in each row but in above matrix R3 doesn’t have pivot element. So,  Inverse of Matrix doesn’t exists.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .