Chapter 10, Problem 1P

Use the rules of matrix multiplication to prove that Eqs. (10.7) and (10.8) follow from Eq. (10.6).

To determine

To prove: The equations $\left[L\right]\left[U\right]=\left[A\right]$ and $\left[L\right]\left\{D\right\}=\left\{B\right\}$ follow the equation $\left[L\right]\left\{\left[U\right]\left\{X\right\}-\left\{D\right\}\right\}=\left[A\right]\left\{X\right\}-\left\{B\right\}$ by the use of matrix multiplication.

Explanation of Solution

Given:

The equations, $\left[L\right]\left[U\right]=\left[A\right]$ and $\left[L\right]\left\{D\right\}=\left\{B\right\}$.

Proof:

Consider a linear equation represented in the matrix form as,

$\left[A\right]\left\{X\right\}=\left\{B\right\}$ …… (1)

where, is A is a regular matrix of size $3\times 3$, X is a variable matrix of size $3\times 1$ and B is the resultant matrix of size $3\times 1$, equation (1) can be rewritten as,

$\left[A\right]\left\{X\right\}-\left\{B\right\}=0$ …… (2)

Suppose equation (2) could be expressed as upper triangular system,

$\left[\begin{array}{ccc}u11& u12& u13\\ 0& u22& u23\\ 0& 0& u33\end{array}\right]\left\{\begin{array}{c}x1\\ x2\\ x3\end{array}\right\}=\left\{\begin{array}{c}d1\\ d2\\ d3\end{array}\right\}$ …… (3)

Equation (3) could be expressed in matrix notation and rearranged as,

$\left[U\right]\left\{X\right\}-\left\{D\right\}=0$ …… (4)

Now assume there is lower diagonal matrix with $1\text{'}s$ on the diagonal,

$\left[L\right]=\left[\begin{array}{ccc}1& 0& 0\\ l21& 1& 0\\ l31& l32& 1\end{array}\right]$

which has property that when equation (4) pre-multiplied by it, equation (2) is the result. That is,

$\left[L\right]\left\{\left[U\right]\left\{X\right\}-\left\{D\right\}\right\}=\left[A\right]\left\{X\right\}-\left\{B\right\}$ .….. (5)

Use distributive law of matrix multiplication $A[B+C]=AB+AC$ for matrix A, B, C,

$\left[L\right]\left\{\left[U\right]\left\{X\right\}\right\}-\left[L\right]\left\{D\right\}=\left[A\right]\left\{X\right\}-\left\{B\right\}$ …… (6)

Use associative law of matrix multiplication $A\left[BC\right]=\left[AB\right]C$ in equation (6) as shown below,

$\left\{\left[L\right]\left[U\right]\right\}\left\{X\right\}-\left[L\right]\left\{D\right\}=\left[A\right]\left\{X\right\}-\left\{B\right\}$ …… (7)

Comparing LHS and RHS of equation (7), this gives,

$\left[L\right]\left[U\right]=\left[A\right]$ and $\left[L\right]\left\{D\right\}=\left\{B\right\}$

Hence, proved that the equations $\left[L\right]\left[U\right]=\left[A\right]$ and $\left[L\right]\left\{D\right\}=\left\{B\right\}$ follow the equation $\left[L\right]\left\{\left[U\right]\left\{X\right\}-\left\{D\right\}\right\}=\left[A\right]\left\{X\right\}-\left\{B\right\}$.