Y = inv(X)
Compute the inverse of a 3-by-3 matrix.
X = [1 0 2; -1 5 0; 0 3 -9]
X = 1 0 2 -1 5 0 0 3 -9
Y = inv(X)
Y = 0.8824 -0.1176 0.1961 0.1765 0.1765 0.0392 0.0588 0.0588 -0.0980
Check the results. Ideally,
Y*X produces the identity matrix. Since
inv performs the matrix inversion using floating-point computations, in practice
Y*X is close to, but not exactly equal to, the identity matrix
ans = 1.0000 0 -0.0000 0 1.0000 -0.0000 0 0 1.0000
Examine why solving a linear system by inverting the matrix using
inv(A)*b is inferior to solving it directly using the backslash operator,
x = A\b.
Create a random matrix
A of order 500 that is constructed so that its condition number,
1e10, and its norm,
1. The exact solution
x is a random vector of length 500, and the right side is
b = A*x. Thus the system of linear equations is badly conditioned, but consistent.
n = 500; Q = orth(randn(n,n)); d = logspace(0,-10,n); A = Q*diag(d)*Q'; x = randn(n,1); b = A*x;
Solve the linear system
A*x = b by inverting the coefficient matrix
toc to get timing information.
tic y = inv(A)*b; t = toc
t = 0.3645
Find the absolute and residual error of the calculation.
err_inv = norm(y-x)
err_inv = 5.0942e-06
res_inv = norm(A*y-b)
res_inv = 5.3472e-07
Now, solve the same linear system using the backslash operator
tic z = A\b; t1 = toc
t1 = 0.3825
err_bs = norm(z-x)
err_bs = 4.7615e-06
res_bs = norm(A*z-b)
res_bs = 3.8876e-15
The backslash calculation is quicker and has less residual error by several orders of magnitude. The fact that
err_bs are both on the order of
1e-6 simply reflects the condition number of the matrix.
The behavior of this example is typical. Using
A\b instead of
inv(A)*b is two to three times faster, and produces residuals on the order of machine accuracy relative to the magnitude of the data.
X is invertible
if there exists a matrix
Y of the same size such
n identity matrix.
Y is called the inverse of
A matrix that has no inverse is singular. A square matrix is singular only when its determinant is exactly zero.
It is seldom necessary to form the explicit inverse
of a matrix. A frequent misuse of
inv arises when
solving the system of linear equations Ax = b.
One way to solve the equation is with
x = inv(A)*b.
A better way, from the standpoint of both execution time and numerical
accuracy, is to use the matrix backslash operator
x = A\b.
This produces the solution using Gaussian elimination, without explicitly
forming the inverse. See
inv performs an LU decomposition of the
input matrix (or an LDL decomposition if the input matrix is Hermitian).
It then uses the results to form a linear system whose solution is
the matrix inverse
inv(X). For sparse inputs,
a sparse identity matrix and uses backslash,