You are an expert in writing code in many different languages. Your goal is to perform code completion for the following code, keeping in mind the rest of the code and the file meta data.
    Metadata details:
    filename: dgtsv.go
    path of file: ./repos/gonum/lapack/gonum
    
    the code of the file until where you have to start completion: // Copyright ©2020 The Gonum Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.

package gonum

import "math"

// Dgtsv solves the equation
//
//	A * X = B
//
// where A is an n×n tridiagonal matrix. It uses Gaussian elimination with
// partial pivoting. The equation Aᵀ * X = B may be solved by swapping the
// arguments for du and dl.
//
// On entry, dl, d and du contain the sub-diagonal, the diagonal and the
// super-diagonal, respectively, of A. On return, the first n-2 elements of dl,
// the first n-1 elements of du and the first n elements of d may 
    
func (impl Implementation) Dgtsv(n, nrhs int, dl, d, du []float64, b []float64, ldb int) (ok bool) {
	switch {
	case n < 0:
		panic(nLT0)
	case nrhs < 0:
		panic(nrhsLT0)
	case ldb < max(1, nrhs):
		panic(badLdB)
	}

	if n == 0 || nrhs == 0 {
		return true
	}

	switch {
	case len(dl) < n-1:
		panic(shortDL)
	case len(d) < n:
		panic(shortD)
	case len(du) < n-1:
		panic(shortDU)
	case len(b) < (n-1)*ldb+nrhs:
		panic(shortB)
	}

	dl = dl[:n-1]
	d = d[:n]
	du = du[:n-1]

	for i := 0; i < n-1; i++ {
		if math.Abs(d[i]) >= math.Abs(dl[i]) {
			// No row interchange required.
			if d[i] == 0 {
				return false
			}
			fact := dl[i] / d[i]
			d[i+1] -= fact * du[i]
			for j := 0; j < nrhs; j++ {
				b[(i+1)*ldb+j] -= fact * b[i*ldb+j]
			}
			dl[i] = 0
		} else {
			// Interchange rows i and i+1.
			fact := d[i] / dl[i]
			d[i] = dl[i]
			tmp := d[i+1]
			d[i+1] = du[i] - fact*tmp
			du[i] = tmp
			if i+1 < n-1 {
				dl[i] = du[i+1]
				du[i+1] = -fact * dl[i]
			}
			for j := 0; j < nrhs; j++ {
				tmp = b[i*ldb+j]
				b[i*ldb+j] = b[(i+1)*ldb+j]
				b[(i+1)*ldb+j] = tmp - fact*b[(i+1)*ldb+j]
			}
		}
	}
	if d[n-1] == 0 {
		return false
	}

	// Back solve with the matrix U from the factorization.
	for j := 0; j < nrhs; j++ {
		b[(n-1)*ldb+j] /= d[n-1]
		if n > 1 {
			b[(n-2)*ldb+j] = (b[(n-2)*ldb+j] - du[n-2]*b[(n-1)*ldb+j]) / d[n-2]
		}
		for i := n - 3; i >= 0; i-- {
			b[i*ldb+j] = (b[i*ldb+j] - du[i]*b[(i+1)*ldb+j] - dl[i]*b[(i+2)*ldb+j]) / d[i]
		}
	}

	return true
}