9.7 Solving systems with inverses  (Page 3/8)

Finding the multiplicative inverse of 3×3 matrices

Unfortunately, we do not have a formula similar to the one for a $2\times 2$ matrix to find the inverse of a $3\times 3$ matrix. Instead, we will augment the original matrix with the identity matrix and use row operations    to obtain the inverse.

Given a $3\times 3$ matrix

augment $A$ with the identity matrix

To begin, we write the augmented matrix    with the identity on the right and $A$ on the left. Performing elementary row operations    so that the identity matrix    appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

Solving a system of linear equations using the inverse of a matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $X$ is the matrix representing the variables of the system, and $B$ is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as

To solve a system of linear equations using an inverse matrix , let $A$ be the coefficient matrix    , let $X$ be the variable matrix, and let $B$ be the constant matrix. Thus, we want to solve a system $AX=B.$ For example, look at the following system of equations.

From this system, the coefficient matrix is

The variable matrix is

And the constant matrix is

Then $AX=B$ looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, $(2^{\mathrm{-1}})2=\left(\frac{1}{2}\right)2=1.$ To solve a single linear equation $ax=b$ for $x,$ we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of $a.$ Thus,