calc: Matrix Tutorial
3.3.2 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]’. 
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
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
3 Matrix Answer 3. (•)
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