Compute the inverse of a 3-by-3 matrix.

X = [1 0 2; -1 5 0; 0 3 -9]

X =3×31 0 2 -1 5 0 0 3 -9

Y = inv(X)

Y =3×30.8824 -0.1176 0.1961 0.1765 0.1765 0.0392 0.0588 0.0588 -0.0980

Check the results. Ideally,Y*Xproduces the identity matrix. Sinceinvperforms the matrix inversion using floating-point computations, in practiceY*Xis close to, but not exactly equal to, the identity matrixeye(size(X)).

Y*X

ans =3×31.0000 0.0000 -0.0000 0 1.0000 -0.0000 0 -0.0000 1.0000

Examine why solving a linear system by inverting the matrix usinginv(A)*bis inferior to solving it directly using the backslash operator,x = A\b.

创建一个随机矩阵Aof order 500 that is constructed so that its condition number,cond(A), is1e10, and its norm,norm(A), is1. The exact solutionxis a random vector of length 500, and the right side isb = 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 systemA*x = bby inverting the coefficient matrixA. Useticandtocto get timing information.

tic y = inv(A)*b; t = toc

t = 0.0302

Find the absolute and residual error of the calculation.

err_inv = norm(y-x)

err_inv = 4.3020e-06

res_inv = norm(A*y-b)

res_inv = 4.1853e-07

Now, solve the same linear system using the backslash operator\.

tic z = A\b; t1 = toc

t1 = 0.0126

err_bs = norm(z-x)

err_bs = 3.1579e-06

res_bs = norm(A*z-b)

res_bs = 2.7938e-15

The backslash calculation is quicker and has less residual error by several orders of magnitude. The fact thaterr_invanderr_bsare both on the order of1e-6simply reflects the condition number of the matrix.

The behavior of this example is typical. UsingA\binstead ofinv(A)*bis two to three times faster, and produces residuals on the order of machine accuracy relative to the magnitude of the data.