Method of Reduction

As the name suggests, in this method the given equations are reduced -- through row transformations -- to a form from which the solution is read off directly, without ever computing a matrix inverse.

The procedure. Start, as in section 2.3, by converting the given linear equations into the matrix equation $AX=B$. Then perform suitable row transformations on $A$, reducing it to an upper triangular, lower triangular, or fully diagonal matrix. The same row transformations, in the same order, are performed simultaneously on the constant matrix $B$. Once this is done, the transformed matrix equation is rewritten back as a system of linear equations -- but now in a form simple enough to solve directly by the elimination method (each equation, taken in the right order, involves one fewer unknown than the one before it).

Worked Example -- two equations, upper triangular. Solve $2x+3y=9$ and $y-x=-2$ by the method of reduction. Writing the equations as $2x+3y=9$ and $-x+y=-2$, the matrix equation is $\begin{bmatrix}2&3\\-1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}9\\-2\end{bmatrix}$. Using $R_2\to 2R_2+R_1$: $\begin{bmatrix}2&3\\0&5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}9\\5\end{bmatrix}$. Rewriting as equations: $2x+3y=9$ …(i) and $5y=5$ …(ii). From (ii), $y=1$; substituting into (i) gives $x=3$. So $x=3,\ y=1$.

Worked Example -- three equations, upper triangular. Solve $x+3y+3z=12$, $x+4y+4z=15$, $x+3y+4z=13$ by the method of reduction. The matrix equation is $\begin{bmatrix}1&3&3\\1&4&4\\1&3&4\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\15\\13\end{bmatrix}$. Using $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$: $\begin{bmatrix}1&3&3\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}12\\3\\1\end{bmatrix}$. This is already upper triangular, so back-substitution gives $z=1$; then $y+z=3\Rightarrow y=2$; then $x+3y+3z=12\Rightarrow x=12-6-3=3$. So $x=3,\ y=2,\ z=1$.

Worked Example -- three equations, lower triangular. Solve $x+y+z=1$, $2x+3y+2z=2$, $x+y+2z=4$. The matrix equation is $\begin{bmatrix}1&1&1\\2&3&2\\1&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}$. Using $R_2\to R_2-R_3$ and $R_1\to R_1-\tfrac12R_3$, then $R_1\to R_1-\tfrac14R_2$, the coefficient matrix reduces to a lower triangular matrix (this time zeros appear above the diagonal instead of below), and the system rewrites -- solved from the top down -- as $x=-2$, then $x+2y=-2\Rightarrow y=0$, then $x+y+2z=4\Rightarrow 2z=6\Rightarrow z=3$. So $x=-2,\ y=0,\ z=3$. …