0.2 Matrices  (Page 8/11)

We write the augmented matrix as follows.

We reduce this matrix using the Gauss-Jordan method.

Multiplying the first row by –2 and adding it to the second row, we get

If we swap the second and third rows, we get

Divide the second row by –2. The result is

Let us do two operations here. 1) Add the second row to first, 2) Add -5 times the second row to the third. And we get

Multiplication of the third row by 2 results in

Multiply the third row by $1/2$ and add it to the second. Also, multiply the third row by $-1/2$ and add it to the first. We get

Therefore, the inverse of matrix $A$ is

One should verify the result by multiplying the two matrices to see if the product does, indeed, equal the identity matrix.

To solve the above equation, we multiply both sides of the equation by the multiplicative inverse of $\frac{2}{3}$ which happens to be $\frac{3}{2}$ . We get

We use the above example as an analogy to show how linear systems of the form $\text{AX}=B$ are solved.

To solve the above equation, first we express the system as

where $A$ is the coefficient matrix, and $B$ is the matrix of constant terms. We get

To solve this system, we multiply both sides of the matrix equation $\text{AX}=B$ by $A^{-1}$ . Since the matrix $A$ is the same matrix $A$ whose inverse we found in [link] ,

Multiplying both sides by $A^{-1}$ , we get

Therefore, $x=2$ , and $y=-3$ .

To solve the above equation, we write the system in the matrix form $\text{AX}=B$ as follows:

$\left[\begin{array}{ccc}1& -1& 1\\ 2& 3& 0\\ 0& -2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 1\\ 5\end{array}\right]$

To solve this system, we need inverse of $A$ . From [link] , we have

$A^{-1}=\left[\begin{array}{ccc}3& -1& -3\\ -2& 1& 2\\ -4& 2& 5\end{array}\right]$

We multiply both sides of the matrix equation $\text{AX}=B$ , by $A^{-1}$ , we get

$\left[\begin{array}{ccc}3& -1& -3\\ -2& 1& 2\\ -4& 2& 5\end{array}\right]\left[\begin{array}{ccc}1& -1& -3\\ 2& 3& 0\\ 0& -2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}3& -1& -3\\ -2& 1& 2\\ -4& 2& 5\end{array}\right]\left[\begin{array}{c}6\\ 1\\ 5\end{array}\right]$

After multiplying the matrices, we get

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right]$