0.2 Matrices  (Page 7/11)

The reduced row-echelon form is given below.

This time the last two equations drop out, and we are left with one equation and three variables. Again, there are infinite number of solutions. But this time the answer must be expressed in terms of two arbitrary constants.

If we let $z=t$ and let $y=s$ , then the first equation $x+\mathrm{2y}-\mathrm{3z}=5$ results in $x=5-\mathrm{2s}+\mathrm{3t}$ .

We rewrite the solution as: $x=5-\mathrm{2s}+\mathrm{3t}$ , $y=s$ , $z=t$ .

The matrices are inverses if the product $\text{AB}$ and $\text{BA}$ both equal $I_2$ , the identity matrix of dimension $2\times 2$ .

Clearly that is the case; therefore, the matrices A and B are inverses of each other.

Suppose $A$ has an inverse, and it is

Then $AB=I$

After multiplying the two matrices on the left side, we get

Equating the corresponding entries, we get four equations with four unknowns as follows:

Solving this system, we get

Therefore, the inverse of the matrix $A$ is

In this problem, finding the inverse of matrix $A$ amounted to solving the system of equations:

Actually, it can be written as two systems, one with variables $a$ and $c$ , and the other with $b$ and $d$ . The augmented matrices for both are given below.

As we look at the two augmented matrices, we notice that the coefficient matrix for both the matrices is the same. Which implies the row operations of the Gauss-Jordan method will also be the same. A great deal of work can be saved if the two right hand columns are grouped together to form one augmented matrix as below.

And solving this system, we get

The matrix on the right side of the vertical line is the $A^{-1}$ matrix.