Matrices
--------

A "matrix" is just a vector of vectors, all the same length.  This
means you can enter a matrix using nested brackets.  You can also use
the semicolon character to enter a matrix.  We'll show both methods
here:

     1:  [ [ 1, 2, 3 ]             1:  [ [ 1, 2, 3 ]
           [ 4, 5, 6 ] ]                 [ 4, 5, 6 ] ]
         .                             .

       [[1 2 3] [4 5 6]]             ' [1 2 3; 4 5 6] RET

We'll be using this matrix again, so type `s 4' to save it now.

Note that semicolons work with incomplete vectors, but they work
better in algebraic entry.  That's why we use the apostrophe in the
second example.

When two matrices are multiplied, the lefthand matrix must have the
same number of columns as the righthand matrix has rows.  Row `i',
column `j' of the result is effectively the dot product of row `i' of
the left matrix by column `j' of the right matrix.

If we try to duplicate this matrix and multiply it by itself, the
dimensions are wrong and the multiplication cannot take place:

     1:  [ [ 1, 2, 3 ]   * [ [ 1, 2, 3 ]
           [ 4, 5, 6 ] ]     [ 4, 5, 6 ] ]
         .

         RET *

Though rather hard to read, this is a formula which shows the product
of two matrices.  The `*' function, having invalid arguments, has been
left in symbolic form.

We can multiply the matrices if we "transpose" one of them first.

     2:  [ [ 1, 2, 3 ]       1:  [ [ 14, 32 ]      1:  [ [ 17, 22, 27 ]
           [ 4, 5, 6 ] ]           [ 32, 77 ] ]          [ 22, 29, 36 ]
     1:  [ [ 1, 4 ]              .                       [ 27, 36, 45 ] ]
           [ 2, 5 ]                                    .
           [ 3, 6 ] ]
         .

         U v t                   *                     U TAB *

Matrix multiplication is not commutative; indeed, switching the order
of the operands can even change the dimensions of the result matrix,
as happened here!

If you multiply a plain vector by a matrix, it is treated as a single
row or column depending on which side of the matrix it is on.  The
result is a plain vector which should also be interpreted as a row or
column as appropriate.

     2:  [ [ 1, 2, 3 ]      1:  [14, 32]
           [ 4, 5, 6 ] ]        .
     1:  [1, 2, 3]
         .

         r 4 r 1                *

Multiplying in the other order wouldn't work because the number of
rows in the matrix is different from the number of elements in the
vector.

(*) *Exercise 1.*  Use `*' to sum along the rows
of the above 2x3 matrix to get `[6, 15]'.  Now use `*' to
sum along the columns to get `[5, 7, 9]'.
See 1: Matrix Answer 1. (*)

An "identity matrix" is a square matrix with ones along the diagonal
and zeros elsewhere.  It has the property that multiplication by an
identity matrix, on the left or on the right, always produces the
original matrix.

     1:  [ [ 1, 2, 3 ]      2:  [ [ 1, 2, 3 ]      1:  [ [ 1, 2, 3 ]
           [ 4, 5, 6 ] ]          [ 4, 5, 6 ] ]          [ 4, 5, 6 ] ]
         .                  1:  [ [ 1, 0, 0 ]          .
                                  [ 0, 1, 0 ]
                                  [ 0, 0, 1 ] ]
                                .

         r 4                    v i 3 RET              *

If a matrix is square, it is often possible to find its "inverse",
that is, a matrix which, when multiplied by the original matrix,
yields an identity matrix.  The `&' (reciprocal) key also computes the
inverse of a matrix.

     1:  [ [ 1, 2, 3 ]      1:  [ [   -2.4,     1.2,   -0.2 ]
           [ 4, 5, 6 ]            [    2.8,    -1.4,    0.4 ]
           [ 7, 6, 0 ] ]          [ -0.73333, 0.53333, -0.2 ] ]
         .                      .

         r 4 r 2 |  s 5         &

The vertical bar `|' "concatenates" numbers, vectors, and matrices
together.  Here we have used it to add a new row onto our matrix to
make it square.

We can multiply these two matrices in either order to get an identity.

     1:  [ [ 1., 0., 0. ]      1:  [ [ 1., 0., 0. ]
           [ 0., 1., 0. ]            [ 0., 1., 0. ]
           [ 0., 0., 1. ] ]          [ 0., 0., 1. ] ]
         .                         .

         M-RET  *                  U TAB *

Matrix inverses are related to systems of linear equations in algebra.
Suppose we had the following set of equations:

         a + 2b + 3c = 6
        4a + 5b + 6c = 2
        7a + 6b      = 3

This can be cast into the matrix equation,

        [ [ 1, 2, 3 ]     [ [ a ]     [ [ 6 ]
          [ 4, 5, 6 ]   *   [ b ]   =   [ 2 ]
          [ 7, 6, 0 ] ]     [ c ] ]     [ 3 ] ]

We can solve this system of equations by multiplying both sides by the
inverse of the matrix.  Calc can do this all in one step:

     2:  [6, 2, 3]          1:  [-12.6, 15.2, -3.93333]
     1:  [ [ 1, 2, 3 ]          .
           [ 4, 5, 6 ]
           [ 7, 6, 0 ] ]
         .

         [6,2,3] r 5            /

The result is the `[a, b, c]' vector that solves the equations.
(Dividing by a square matrix is equivalent to multiplying by its
inverse.)

Let's verify this solution:

     2:  [ [ 1, 2, 3 ]                1:  [6., 2., 3.]
           [ 4, 5, 6 ]                    .
           [ 7, 6, 0 ] ]
     1:  [-12.6, 15.2, -3.93333]
         .

         r 5  TAB                         *

Note that we had to be careful about the order in which we multiplied
the matrix and vector.  If we multiplied in the other order, Calc
would assume the vector was a row vector in order to make the
dimensions come out right, and the answer would be incorrect.  If you
don't feel safe letting Calc take either interpretation of your
vectors, use explicit Nx1 or 1xN matrices instead.  In this case, you
would enter the original column vector as `[[6], [2], [3]]' or `[6; 2;
3]'.

(*) *Exercise 2.* Algebraic entry allows you to make vectors and
matrices that include variables.  Solve the following system of
equations to get expressions for `x' and `y' in terms of `a' and `b'.

        x + a y = 6
        x + b y = 10

See 2: Matrix Answer 2. (*)

(*) *Exercise 3.* A system of equations is "over-determined" if it has
more equations than variables.  It is often the case that there are no
values for the variables that will satisfy all the equations at once,
but it is still useful to find a set of values which "nearly" satisfy
all the equations.  In terms of matrix equations, you can't solve `A X
= B' directly because the matrix `A' is not square for an
over-determined system.  Matrix inversion works only for square
matrices.  One common trick is to multiply both sides on the left by
the transpose of `A': `trn(A)*A*X = trn(A)*B'.  Now `trn(A)*A' is a
square matrix so a solution is possible.  It turns out that the `X'
vector you compute in this way will be a "least-squares" solution,
which can be regarded as the "closest" solution to the set of
equations.  Use Calc to solve the following over-determined system:

         a + 2b + 3c = 6
        4a + 5b + 6c = 2
        7a + 6b      = 3
        2a + 4b + 6c = 11

See 3: Matrix Answer 3. (*)