Find the inverse of the given matrix

[2 4 3]

[3 -4 -4]

[5 0 -1]

Can you please teach me the steps of sloving the inverse of this matrix? Thank you very much.

Printable View

- May 20th 2008, 08:12 PMalgebramatrix inverse
Find the inverse of the given matrix

[2 4 3]

[3 -4 -4]

[5 0 -1]

Can you please teach me the steps of sloving the inverse of this matrix? Thank you very much. - May 20th 2008, 08:28 PMo_O
Which methods have you learned in class? You can perform row operations on your matrix until you get it into row echelon form and perform the exact same operations on the identity matrix to get your inverse.

Or you can utilize determinants ... Any method in particular? - May 20th 2008, 09:00 PMalgebra
Hi o_O,

Thank you very much for your reply.

I am just in the beginning of linear algebra.

Here is the method that we used :

1) form the augmented matrix [A l I sub n] and Use elementary row operations.

I saw the answer of this question in the book is " does not exist" . I don't know how to reach this answer from the given question. - May 20th 2008, 09:44 PMo_O
$\displaystyle \left[ \begin{array}{ccc}2 & 4 & 3 \\ 3 & -4 & -4 \\ 5 & 0 & -1 \end{array} \Bigg| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]$

Now, I presume you understand that whatever operations you do to reduce A to the identity matrix, you will get A-inverse if you perform those exact same operations to the identity matrix.

In order for an inverse to not exist, a row of 0's must occur in your augmented matrix after performing some row operations as this will prevent any possibilities of achieving your identity matrix. For example, [0 0 0] -> [0 0 1] you can't achieve this via elementary row operations.

If we just focus on trying to reduce A to the identity matrix, we will wind up with exactly that - a row of zeroes:

$\displaystyle \left[ \begin{array}{ccc}2 & 4 & 3 \\ 3 & -4 & -4 \\ 5 & 0 & -1 \end{array}\right] $$\displaystyle {\color{white}.} \quad \stackrel{R_{1} \iff R_{3}}{\longrightarrow}$$\displaystyle {\color{white}.} \quad \left[ \begin{array}{ccc}5 & 0 & -1 \\ 3 & -4 & -4 \\ 2 & 4 & 3 \end{array}\right]$$\displaystyle {\color{white}.} \quad \stackrel{R_{3}' = 3R_{3} -2R_{2}}{\longrightarrow}$$\displaystyle {\color{white}.} \quad \left[ \begin{array}{ccc}5 & 0 & -1 \\ 3 & -4 & -4 \\ 0 & 20 & 17 \end{array}\right]$

$\displaystyle \stackrel{R_{2}' = 5R_{2} - 3R_{1}}{\longrightarrow} \left[ \begin{array}{ccc}5 & 0 & -1 \\ 0 & -20 & -17 \\ 0 & 20 & 17 \end{array}\right]$

Now whatever you do with this last matrix, you will wind up with a row of 0's and will thus have no inverse. You can see more here (scroll down to Example 3 for a similar example) - May 20th 2008, 10:36 PMMoo
Hello,

Or you can just say that the determinant is 0, therefore the matrix has no inverse. - May 21st 2008, 11:27 AMo_O
Of course but the OP was asked to determine the matrix's invertibility by means of elementary row operations.