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How to find the inverse of the matrix Question number 19

How To Find The Inverse Of The Matrix Question Number 19 class=

Sagot :

Okay, here we have this:

We need to find the inverse of the matrix, let's do it:

[tex]\begin{bmatrix}{2} & {4} & {1} \\ {-1} & {1} & {-1} \\ {1} & {4} & {0}\end{bmatrix}[/tex]

For that we are going to make the augmented form with the identity matrix and convert the original matrix into the identity:

[tex]\begin{gathered} \begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ -1 & 1 & -1 & | & 0 & 1 & 0 \\ 1 & 4 & 0 & | & 0 & 0 & 1\end{pmatrix} \\ =\begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ 0 & 3 & -\frac{1}{2} & | & \frac{1}{2} & 1 & 0 \\ 1 & 4 & 0 & | & 0 & 0 & 1\end{pmatrix}\text{ }R_2\leftarrow R_2+\frac{1}{2}R_1 \\ =\begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ 0 & 3 & -\frac{1}{2} & | & \frac{1}{2} & 1 & 0 \\ 0 & 2 & -\frac{1}{2} & | & -\frac{1}{2} & 0 & 1\end{pmatrix}\text{ }R_3\leftarrow R_3-\frac{1}{2}R_1 \\ =\begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ 0 & 3 & -\frac{1}{2} & | & \frac{1}{2} & 1 & 0 \\ 0 & 0 & -\frac{1}{6} & | & -\frac{5}{6} & -\frac{2}{3} & 1\end{pmatrix}R_3\leftarrow R_3-2/3R_2 \\ =\begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ 0 & 3 & -\frac{1}{2} & | & \frac{1}{2} & 1 & 0 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_3\leftarrow-6R_3 \\ =\begin{pmatrix}2 & 4 & 1 & | & 1 & 0 & 0 \\ 0 & 3 & 0 & | & 3 & 3 & -3 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_2\leftarrow R_2+\frac{1}{2}R_3 \\ =\begin{pmatrix}2 & 4 & 0 & | & -4 & -4 & 6 \\ 0 & 3 & 0 & | & 3 & 3 & -3 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_1\leftarrow R_1-R_3 \\ =\begin{pmatrix}2 & 4 & 0 & | & -4 & -4 & 6 \\ 0 & 1 & 0 & | & 1 & 1 & -1 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_2\leftarrow\frac{1}{3}R_2 \\ =\begin{pmatrix}2 & 0 & 0 & | & -8 & -8 & 10 \\ 0 & 1 & 0 & | & 1 & 1 & -1 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_1\leftarrow R_1-4R_2 \\ =\begin{pmatrix}1 & 0 & 0 & | & -4 & -4 & 5 \\ 0 & 1 & 0 & | & 1 & 1 & -1 \\ 0 & 0 & 1 & | & 5 & 4 & -6\end{pmatrix}R_1\leftarrow\frac{1}{2}R_1 \end{gathered}[/tex]

Finally the inverse is on the right side of the augmented matrix:

[tex]=\begin{pmatrix}-4 & -4 & 5 \\ 1 & 1 & -1 \\ 5 & 4 & -6\end{pmatrix}[/tex]

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